# 60+ Solved Speed, Velocity, and Acceleration Problems for High School

## Speed, velocity, and acceleration problems for a straight-line path:

Over 60 simple problems on speed, velocity, and acceleration with descriptive solutions are presented for high school students. In each solution, you can find a brief tutorial. These motion problems update regularly.

### Speed and velocity Problems:

Problem (1): What is the speed of a rocket that travels $8000\,{\rm m}$ in $13\,{\rm s}$?

Solution: Speed is the total distance divided by the elapsed time so the rocket's speed is $\frac{8000}{13}=615.38\,{\rm m/s}$.

Problem (2): How long will it take if you travel $400\,{\rm km}$ with an average speed of $100\,{\rm m/s}$?

Solution: Average speed is the ratio of total distance to the total time. Thus, the elapsed time is \begin{align*} t&=\frac{\text{total distance}}{\text{average speed}}\\&=\frac{400\times 10^{3}\,{\rm m}}{100\,{\rm m/s}}\\&=4000\,{\rm s}\end{align*} To convert it to hours it must be divided by $3600\,{\rm s}$ which get $t=1.11\,{\rm h}$.

Problem (3): A person walks $100\,{\rm m}$ in $5$ minutes, then $200\,{\rm m}$ in $7$ minutes and finally $50\,{\rm m}$ in $4$ minutes. Find its average speed?

Solution: First find its total distance traveled $D$ by summing all distances in each section which gets $D=100+200+50=350\,{\rm m}$. Now by definition of average speed, divide it by the total time elapsed $T=5+7+4=11$ minutes. But keep in mind that since distance is in SI units so the time traveled must also be in SI units which is $m/s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\&=\frac{350\,{\rm m}}{11\times 60\,{\rm s}}\\&=0.53\,{\rm m/s}\end{align*}

Problem (4): A person walks $750\,{\rm m}$ due north, then $250\,{\rm m}$ due east. If the entire walk takes $12$ minutes, find the person's average velocity?

Solution: Average velocity is displacement divided by the time elapsed. Displacement is also a vector which obeys the addition vector rules. Thus, in this problem, add each displacement to get the total displacement

In the first part, displacement is $\Delta x_1=750\,\hat{j}$ and for the second part $\Delta x_2=250\,\hat{i}$. The total displacement vector is $\Delta x=\Delta x_1+\Delta x_2=750\,\hat{i}+250\,\hat{j}$ with magnitude of  \begin{align*}|\Delta x|&=\sqrt{(750)^{2}+(250)^{2}}\\&=790.5\,{\rm m}\end{align*}Therefore, the magnitude of the average velocity is $v=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}$

Problem (5): An object moves along a straight line. First it travels at a velocity of $12\,{\rm m/s}$ for $5\,{\rm s}$ and then continues at the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?

Solution: Average velocity is displacement divided by the elapsed time i.e. $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$. In this problem, object goes through two stages with two different displacements so add them to find the total displacement. Thus,$\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}$Displacements are obtained as $x_1=v_1\,t_1=12\times 5=60\,{\rm m}$ and $x_2=v_2\,t_2=20\times 3=60\,{\rm m}$. Therefore, we have \begin{align*} \bar{v}&=\frac{x_1+x_2}{t_1+t_2}\\&=\frac{60+60}{5+3}\\&=15\,{\rm m/s}\end{align*}

Problem (6): A plane flies the distance between two cities in $1$ hour and $30$ minutes with a velocity of $900\,{\rm km/h}$. Another plane covers that distance with $600\,{\rm km/h}$. What is the flight time of the second plane?

Solution: first find the distance between two cities as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\&=\frac{1350}{600}\\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour which converts in minutes as $2$ hours and ($0.25\times 60=15$) minutes.

Problem (7): A particle is moving along a straight-line path. At times $t_1=2\,{\rm s}$ and $t_2=4\,{\rm s}$ its position from the origin is $x_1=2\,{\rm m}$ and $x_2=-8\,{\rm m}$. Find average velocity and average speed of the particle.

Solution: Average speed defines as the ratio of the path length (distance) to the total elapsed time, $\text{Average speed} = \frac{\text{path length}}{\text{elapsed time}}$ On the other hand, average velocity is the displacement $\Delta x=x_2-x_1$ divided by the elapsed time $\Delta t$.

The distance between those two points is $D=12\,{\rm m}$ but its displacement is $\Delta x=x_2-x_1=-8-4=-12\,{\rm m}$. Distance is a scalar quantity and its value is always positive but displacement is a vector. In above the minus sign of the displacement indicates its direction which is toward the $-x$ axis.

Thus, average speed is $=\frac{12}{4-2}=6\,{\rm m/s}$ and average velocity is $\bar{v}=\frac{-12}{4-2}=-6\,{\rm m/s}$. The minus signs shows the direction of the velocity which is in the same direction of displacement.

Problem (8): An object moving along the $x$-axis. At the instant $t=1\,{\rm s}$, it is at the position $x=+4\,{\rm m}$ and has a velocity of $4\,{\rm m/s}$. At $t=5\,{\rm s}$, the object is at the location $x=+9\,{\rm m}$ and its velocity is $-12\,{\rm m/s}$. What is its average acceleration during the time interval $1\leq t\leq 5$?

Solution: Average acceleration is defined as the difference in velocities divided by the time interval that change is occurred. Thus we have\begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\&=\frac{v_2-v_1}{t_2-t_1}\\&=\frac{-12-4}{5-1}\\&=-4\,{\rm m/s^2}\end{align*} the negative indicates that the direction of the average acceleration vector is toward the $-x$ axis.

Problem (9): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity $30\,{\rm m/s}$ and finally $4\,{\rm s}$ with an average velocity $25\,{\rm m/s}$. What is its average velocity across the whole path?

Solution: There are three different parts with different average velocities. Displacements in the corresponding parts are obtained as \begin{align*}\Delta x_1&=v_1\,\Delta t_1\\&=10\times 4=40\,{\rm m}\\\Delta x_2&=v_2\,\Delta t_2\\&=30\times 2=60\,{\rm m}\\\Delta x_3&=v_3\,\Delta t_3\\&=25\times 4=100\,{\rm m}\end{align*}Now use the definition of average velocity $\bar{v}=\frac{\Delta x_{tot}}{\Delta t_{tot}}$ to find it over the whole path\begin{align*}\bar{v}&=\frac{\Delta x_{tot}}{\Delta t_{tot}}\\&=\frac{\Delta x_1+\Delta x_2+\Delta x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\&=\frac{40+60+100}{4+2+4}\\&=20\,{\rm m/s}\end{align*}

Problem (10): An object moving along a straight-line path. It travels with an average velocity $2\,{\rm m/s}$ for $20\,{\rm s}$ and $12\,{\rm m/s}$ for $t$ seconds. If the total average velocity across the whole path is $10\,{\rm m/s}$, then find the unknown time $t$?

Solution: In this problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1+\Delta x_2$ with different average velocities and times elapsed, so the total average velocity across the whole path is obtained as \begin{align*}\bar{v}&=\frac{\Delta x}{\Delta t}\\&=\frac{\Delta x_1+\Delta x_2}{\Delta t_1+\Delta t_2}\\&=\frac{\bar{v}_1\,t_1+\bar{v}_2\,t_2}{t_1+t_2}\\10&=\frac{2\times 20+12\times t}{20+t}\\\Rightarrow t&=80\,{\rm s}\end{align*}

Note: whenever a moving object, covers distances $x_1,x_2,x_3,\cdots$ in $t_1,t_2,t_3,\cdots$ with constant or average velocities $v_1,v_2,v_3,\cdots$ along a straight-line without changing its direction, then its total average velocity across the whole path is obtained by one of the following formulas

• Distances and times are known:$\bar{v}=\frac{x_1+x_2+x_3+\cdots}{t_1+t_2+t_3+\cdots}$
• Velocities and times are known: $\bar{v}=\frac{v_1\,t_1+v_2\,t_2+v_3\,t_3+\cdots}{t_1+t_2+t_3+\cdots}$
• Distances and velocities are known:$\bar{v}=\frac{x_1+x_2+x_3+\cdots}{\frac{x_1}{v_1}+\frac{x_2}{v_2}+\frac{x_3}{v_3}+\cdots}$

Problem (11): A car travels one-fourth of its path with a constant velocity of $10\,{\rm m/s}$, and the remaining with a constant velocity of $v_2$. If the total average velocity across the whole path is $16\,{\rm m/s}$, then find the $v_2$?

Solution: This is the third case of the preceding note. Let the length of the path be $L$ so \begin{align*}\bar{v}&=\frac{x_1+x_2}{\frac{x_1}{v_1}+\frac{x_2}{v_2}}\\16&=\frac{\frac 14\,L+\frac 34\,L}{\frac{\frac 14\,L}{10}+\frac{\frac 34\,L}{v_2}}\\\Rightarrow v_2&=20\,{\rm m/s}\end{align*}

Problem (12): An object moves along a straight-line path. It travels for $t_1$ seconds with an average velocity $50\,{\rm m/s}$ and $t_2$ seconds with constant velocity $25\,{\rm m/s}$. If the total average velocity across the whole path is $30\,{\rm m/s}$, then find the ratio $\frac{t_2}{t_1}$?

Solution: the velocities and times are known so we have \begin{align*}\bar{v}&=\frac{v_1\,t_1+v_2\,t_2}{t_1+t_2}\\30&=\frac{50\,t_1+25\,t_2}{t_1+t_2}\\\Rightarrow \frac{t_2}{t_1}&=4\end{align*}

### Acceleration Problems:

Problem (13): A car moving with $15\,{\rm m/s}$ uniformly slows its velocity. It comes to a complete stop in $10\,{\rm s}$. What is its acceleration?

Solution: Average acceleration is difference in velocities divided by the time taken so we have\begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\&=\frac{v_2-v_1}{\Delta t}\\&=\frac{0-15}{10}\\&=-1.5\,{\rm m/s^2}\end{align*}The minus sign indicates the direction of the acceleration vector which is toward the $-x$ direction.

Problem (14): An object moves with constant acceleration along a straight line. If its velocity at instant of $t_1 = 3\,{\rm s}$ is $10\,{\rm m/s}$ and at the moment of $t_2 = 8\,{\rm s}$ is $20\,{\rm m/s}$, then what is its initial speed?

Solution: Let the initial speed at time $t=0$ be $v_0$. Now apply average acceleration definition in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ and equate them.\begin{align*}\text{average acceleration}\ \bar{a}&=\frac{\Delta v}{\Delta t}\\\frac{v_1 - v_0}{t_1-t_0}&=\frac{v_2-v_0}{t_2-t_0}\\ \frac{10-v_0}{3-0}&=\frac{20-v_0}{8-0}\\\Rightarrow v_0 &=4\,{\rm m/s}\end{align*} In the above, $v_1$ and $v_2$ are the velocities at moments $t_1$ and $t_2$, respectively.

Problem (15): For $10\,{\rm s}$, the velocity of a car which travels with a constant acceleration, changes from $10\,{\rm m/s}$ to $30\,{\rm m/s}$. How far does the car travel?

Solution: Known: $\Delta t=10\,{\rm s}$, $v_1=10\,{\rm m/s}$ and $v_2=30\,{\rm m/s}$.

Method (I) Without computing the acceleration: Recall that in the case of constant acceleration, we have the following kinematic equations for average velocity and displacement:\begin{align*}\text{average velocity}:\,\bar{v}&=\frac{v_1+v_2}{2}\\\text{displacement}:\,\Delta x&=\frac{v_1+v_2}{2}\times \Delta t\\\end{align*}where $v_1$ and $v_2$ are the velocities in a given time interval. Now we have \begin{align*} \Delta x&=\frac{v_1+v_2}{2}\\&=\frac{10+30}{2}\times 10\\&=200\,{\rm m}\end{align*}

Method (II) with computing acceleration: Using the definition of average acceleration, first determine it as below \begin{align*}\bar{a}&=\frac{\Delta v}{\Delta t}\\&=\frac{30-10}{10}\\&=2\,{\rm m/s^2}\end{align*} Since the velocities at the initial and final points of the problem are given so use the below time-independent kinematic equation to find the required displacement \begin{align*} v_2^{2}-v_1^{2}&=2\,a\Delta x\\(30)^{2}-(10)^{2}&=2(2)\,\Delta x\\\Rightarrow \Delta x&=200\,{\rm m}\end{align*}

Problem (16): A car moves from rest to a speed of $72\,{\rm km/h}$ in $4\,{\rm s}$. Find the acceleration of the car?

Solution: Known: $v_1=0$, $v_2=72\,{\rm km/h}$, $\Delta t=3\,{\rm s}$.
Average acceleration defined as difference in velocities divided by the time interval between those points \begin{align*}\bar{a}&=\frac{v_2-v_1}{t_2-t_1}\\&=\frac{20-0}{4}\\&=5\,{\rm m/s^2}\end{align*}
In above, we converted the $km/h$ to the SI unit of velocity ($m/s$) as $1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\,m/s$ so we get
$72\,km/h=72\times \frac{10}{36}=20\,m/s$.

Problem (17): A race car accelerate from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity $v_f = 30\,{\rm m/s}$ in a time interval $2\,{\rm s}$. Determine its average acceleration?

Solution:
Known: $v_i=10\,{\rm m/s}$,$v_f = 30\,{\rm m/s}$,$\Delta t=2\,{\rm s}$.
Applying definition of average acceleration, we get \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{30-10}{2}\\&=10\,{\rm m/s^2}\end{align*}

Problem (18): A motorcycle starts its trip along a straight line with a velocity of $10\,{\rm m/s}$ and ends with $20\,{\rm m/s}$ in the opposite direction in a time interval of $2\,{\rm s}$. What is the average acceleration of the car?

Solution:
Known: $v_i=10\,{\rm m/s}$ ,$v_f=20\,{\rm m/s}$, $\Delta t=2\,{\rm s}$, $\bar{a}=?$.
Using average acceleration definition we have \begin{align*}\bar{a}&=\frac{v_f-v_i}{\Delta t}\\&=\frac{(-20)-10}{2}\\&=-15\,{\rm m/s^2}\end{align*}Recall that in the definition above, velocities are vector quantities. Final velocity is in the opposite direction with the initial velocity so a negative must be included.

Problem (19): A car travels along a straight line with a uniform acceleration. If its velocity at the instant of $t_1=2\,{\rm s}$ is $36\,{\rm km/s}$ and at the moment $t_2=6\,{\rm s}$ is $72\,{\rm km/h}$, then find its initial velocity (at $t_0=0$)?

Solution: Use the equality of definition of average acceleration $a=\frac{v_f-v_i}{t_f-t_i}$ in the time intervals $[t_0,t_1]$ and $[t_0,t_2]$ to find the initial velocity as below \begin{align*}\frac{v_2-v_0}{t_2-t_0}&=\frac{v_1-v_0}{t_1-t_0}\\ \frac{20-v_0}{6-0}&=\frac{10-v_0}{2-0}\\\Rightarrow v_0&=5\,{\rm m/s}\end{align*}

### One-Dimensional Motion Problems with Constant Acceleration

Problem (20): An object moves the distance of $45\,{\rm m}$ in the time interval $5\,{\rm s}$ with an initial velocity and acceleration of $v_0$ and $2\,{\rm m/s^2}$, respectively. What is the initial velocity $v_0$?

Solution: Known: $\Delta x=45\,{\rm m}$, $\Delta t=5\,{\rm s}$, $a=2\,{\rm m/s^2}$, $v_0=?$.
In this motion problem, use the following kinematic equation to find the unknown initial velocity \begin{align*}\Delta x&=\frac 12\,at^{2}+v_0 t\\&=\frac 12 (2)(5)^{2}+v_0 (5)\\&\Rightarrow v_0\\&=4\,{\rm m/s}\\ \end{align*}

Problem (21): An object, without change in direction, travels a distance of $50\,{\rm m}$ with an initial speed $5\,{\rm m/s}$ in $4\,{\rm s}$. Find the object's velocity at the end of the given time interval?

Solution:
Known: $\Delta x= 50\,{\rm m}$, $v_i=5\,{\rm m/s}$, $\Delta t=4\,{\rm s}$, $v_f=?$
With the above known values, we only use the following displacement kinematic equation to first find the acceleration \begin{align*} \Delta x&=\frac 12\,at^{2}+v_i\,t\\50&=\frac 12 (a)(4)^{2}+(5)(4)\\\Rightarrow a&=\frac{30}{8}=\frac{15}{4}\end{align*} Now apply the below kinematic formula to find the final velocity \begin{align*}v_f&=v_i+a\,t\\&=5+\frac{15}{4}\times 4=20\,{\rm m/s}\end{align*}
Alternative solution: Since in this problem we have two unknowns that is acceleration and final velocity and the motion is constant acceleration, so one can use the below total displacement formula \begin{align*}\Delta x&=\frac{v_i+v_f}{2}\times \Delta t\\50&=\frac{5+v_f}{2}\times (4)\\\Rightarrow v_f&=20\,{\rm m/s}\end{align*}

Problem (22): A car starts its motion from rest with a constant acceleration of $4\,{\rm m/s^2}$. What is the average velocity of the car in the first $5\,{\rm s}$ of the motion?

Solution: Recall that once you have initial and final velocities of a moving object during a constant acceleration motion, then you can use $\bar{v}=\frac{v_i+v_f}2$ to find the average acceleration. In this problem, $v_i=0$ and final velocity is obtained as \begin{align*}v_f&=v_0+a\,t\\&=0+(4)(5)=20\,{\rm m/s}\end{align*} Now use the above formula to find the average velocity as \begin{align*}\bar{v}&=\frac{0+20}{2}\\&=10\,{\rm m/s}\end{align*}

Problem (23): A particle moves from rest with uniform acceleration and travels $40\,{\rm m}$ in $4\,{\rm s}$. At what distance from the origin is this particle at the instant of $t=10\,{\rm s}$?

Solution:
Known: $\Delta x=40\,{\rm m}$, $\Delta t_1=t-1-t_0=4\,{\rm s}$,$\Delta t_2=t-2-t_0=10\,{\rm s}$
First, use the displacement kinematic equation to find the acceleration as \begin{align*}\Delta x&=\frac 12 a\,t^{2}+v_0 t\\ 40&=\frac 12 (a)(4)^{2}+0\\\Rightarrow a&=5\,{\rm m/s^2}\end{align*} Now use again that formula to find the displacement at the moment $t=10\,{\rm s}$. \begin{align*} \Delta x&=\frac 12 a\,t^{2}+v_0 t\\&=\frac 12 (5)(10)^{2}+0\\&=250\,{\rm m}\end{align*}

Problem (24): An object starts its trip from rest with a constant acceleration. At instant $t=2\,{\rm s}$ is $1$ meter away from origin and at $t=4\,{\rm s}$ is $13\,{\rm m}$ away. At the moment of starting the motion, the object was at what distance away from the origin?

Solution:
Known: $v_0=0$, $t_1=2\,{\rm s}$, $x_1=1\,{\rm m}$,$t_2=4\,{\rm s}$, $x_2=13\,{\rm m}$, $t_0=0$ and $x_0=?$
Substitute the known values into the kinematic equation $x=\frac 12 a\,t^{2}+v_0t+x_0$ which gives two equations with two unknowns \begin{align*}x&=\frac 12 a\,t^{2}+v_0t+x_0\\1&=\frac 12 a\,(2)^{2}+x_0\\13&=\frac 12 a\,(4)^{2}+x_0\end{align*} Multiply the first equation by $-1$ and sum with thee second equation gives $a=2\,{\rm m/s^{2}}$ and $x_0=-3\,{\rm m}$.

Problem (25): A car moves at a speed of $72\,{\rm km/h}$ along a straight path. The driver suddenly brakes and the car comes to a complete stop after $5\,{\rm s}$. Suppose that during the decelerating period, the car's acceleration remains constant. How far has the car traveled between applying the brake and come to rest?

Solution: In all kinematic problems, you must first identify two points with known kinematic variables (i.e. $x,v,a$) and then apply equations between those points.

In this problem, the moment of braking, the car's velocity is known which can be choose as the initial point with initial velocity $72\,{\rm km/h}$. The other point is the end of the path with $v_f=0$. Using kinematic formula $v_f=v_i+at$ one can find the car's acceleration as \begin{align*} v_f&=v_i+at\\0&=20+(a)(5)\\\Rightarrow a&=-4\,{\rm m/s^2}\end{align*} Now apply the kinetic formula below to find the total displacement between braking and resting points \begin{align*}v_f^{2}-v_i^{2}&=2a\Delta x\\0-(20)^{2}&=2(-4)\Delta x\\\Rightarrow \Delta x&=50\,{\rm m}\end{align*}
Alternative Solution: Between the above points we can apply the well-known kinematic equation below to find total displacement \begin{align*}\Delta x&=\frac{v_i+v_f}{2}\,t\\&=\frac{0+20}{2}\times 5\\&=50\,{\rm m}\end{align*}

Problem (26): A motorcycle starts its trip along a straight path from position $x_0=5\,{\rm m}$ with a speed of $8\,{\rm m/s}$ at a constant rate. At position $x=8.5\,{\rm m}$, its speed is $6\,{\rm m/s}$. Find the displacement equation of this motion as a function of time.

Solution: The displacement as a function of time is given by $x=\frac 12 at^{2}+v_0 t+x_0$ where $x_0$ is the initial position at time $t_0=0$. We must apply kinematic equations on two arbitrary points with known date which in this case are: $v_0=8\,{\rm m/s}$, $v_f=6\,{\rm m/s}$.

Using the kinematic formula $v_f^{2}-v_i^{2}=2a\,\Delta x$, one can find the unknown acceleration. Before any computing, we see that the speed is decreasing so a negative acceleration must be obtained. \begin{align*}v_f^{2}-v_i^{2}&=2a\,\underbrace{x_2-x_1}_{\Delta x}\\(6)^{2}-(8)^{2}&=2\,a\,(8.5-5)\\-28&=7\,a\\\Rightarrow a&=-4\,{\rm m/s^2}\end{align*} Now put the known values into the displacement formula to find its time-dependence \begin{align*}x&=\frac 12 at^{2}+v_0 t+x_0\\&=\frac 12 (-4)t^{2}+8t+5\\\Rightarrow x&=-2t^{2}+8t+5\end{align*}

Problem (27): Two cars start racing to reach a same destination at speeds of $54\,{\rm km/h}$ and $108\,{\rm km/h}$. If the faster car reaches two hours earlier, What is the distance between origin to destination?

Solution: Let the slower car be $v_B=54\,{\rm km/h}$ with a total time $t$ for covering the total path $D$. Two hours earlier for faster car, say $v_A=108\,{\rm km/h}$ means $t-2$. The distance traveled by $A$ and $B$ are the same i.e. $D_A=D_B$ so using the definition of average velocity we have \begin{align*}v_A\,t_A&=v_B\,t_B\\108\times (t-2)&=54\times t\\\Rightarrow t&=4\,{\rm h}\end{align*} Now substitute it for one of the cars as $D_A=v_A\,t_A=108\times(4-2)=216\,{\rm m}$ to find the total distance between origin to destination.

Problem (28): A particle starts moving with a constant acceleration $4\,{\rm m/s^2}$ from rest along a straight line. $2\,{\rm s}$ after starting, it decelerates its motion and comes to a complete stop at the moment of $t=4\,{\rm s}$. What is the total distance traveled by this moving object?

Solution: This problem consists of two parts, one has acceleration $a_1$ (for two seconds) and the other deceleration $a_2$ (for two seconds). The displacement to where deceleration starts is calculated as \begin{align*}\Delta x_1&=\frac 12 a_1\,t^{2}+v_0\,t\\&=\frac 12 (4)(2)^{2}+0\\&=8\,{\rm m}\end{align*}The velocity at the starting point of deceleration is determined as \begin{align*}v_f&=v_i+a_1\,t\\&=0+(4)(2)\\&=8\,{\rm m/s}\end{align*}The velocity at the and of the path is also zero (come to a complete rest) so we have \begin{align*}v_f&=v_i+a\,t\\0&=8+a_2\,(2)\\\Rightarrow a_2&=-4\,{\rm m/s}\end{align*}Now you can find the displacement for the deceleration part as \begin{align*}\Delta x_2&=\frac 12\,a_2\,t^{2}+v_0\,t\\&=\frac 12\,(-4)(2)^{2}+(8)(2)\\&=8\,{\rm m}\end{align*}Therefore, the total displacement is $D=\Delta x_1+\Delta x_2=16\,{\rm m}$.

Problem (29): An object moving with a slowing acceleration along a straight line. After $10\,{\rm s}$ and covering distance $60\,{\rm m}$, its velocity reaches $4\,{\rm m/s}$. What is its initial velocity?

Solution: Apply the acceleration-independent kinematic equation $\Delta x=\frac{v_i+v_f}2\,\Delta t$ between those points with distance $60\,{\rm m}$. \begin{align*}\Delta x&=\frac{v_i+v_f}2\,\Delta t\\60&=\frac{v_i+4}2\,(10)\\\Rightarrow v_i&=8\,{\rm m/s}\end{align*}

Practice Problem (30): A bus starts moving from rest along a straight line with a constant acceleration of $2\,{\rm m/s^2}$. After some time its motion becomes uniform and finally comes to rest with an acceleration of $1\,{\rm m/s^2}$. The total displacement covered by the bus is $175\,{\rm m}$, and $10\,{\rm s}$ takes time to complete the uniform motion. What is its average velocity across the whole path?

Solution: this left for you as a practice problem.

Problem (31): The position of an object as a function of time is given by $x=\frac{t^{3}}{3}+2t^{2}+4t$. What is its average acceleration in the time interval $1\,{\rm s}$ and $3\,{\rm s}$?

Solution: Average acceleration is defined as difference in velocities divided by the time interval $\bar{a}=\frac{\Delta v}{\Delta t}$. In this problem the position-time equation given so by differentiating find its velocity as \begin{align*}v&=\frac {d\,x}{dt}\\&=\frac {d}{dt}\left(\frac{t^{3}}{3}+2t^{2}+4t\right)\\&=t^{2}+4t+4\end{align*} Now compute the velocities at the given instants as \begin{align*}v(t=1)&=(1)^{2}+4(1)+4=9\,{\rm m/s}\\v(t=3)&=(3)^{2}+4(3)+4=25\,{\rm m/s}\\\Delta v&=25-9=16\,{\rm m/s}\end{align*}Therefore, the average acceleration is determined as $\bar{a}=\frac {16}{2}=8\,{\rm m/s^{2}}$.

Practice Problem (32): An object starts moving from rest with an acceleration of $a$. After $t$ seconds it brakes and comes to a stop with an acceleration of $2a$. If the total displacement over the whole time interval is $60\,{\rm m}$, What is the displacement in the first $t$-seconds?

Solution: Solve yourself.

Problem (33): The position - time equations of two moving objects along $x$-axis is as follows: $x_1=2t^{2}-8t$ and $x_2=-2t^{2}+4t-14$. These two objects how many times meet each other in the time interval $t=0$ through $t=5\,{\rm s}$?

Solution: once the position equations of two objects are given, equating those equations and solving for $t$, you can find the time when they reach each other. In this problem, we have\begin{align*} x_1&=x_2\\ 2t^{2}-8t&=-2t^{2}+4t-14\end{align*} Rearranging above, we get $4t^{2}-12t+14=0$. Applying quadratic formula yield a negative discriminant $(b^{2}-4\,a\,c)<0$ which means there is no solution for this equation. Thus, those objects never meet each other.

Problem (34): An object starts moving from rest from position $x_0=4\,{\rm m}$ with an initial velocity $4\,{\rm m/s}$ and constant acceleration. At position $x=10\,{\rm m}$ its velocity is $8\,{\rm m/s}$. Find its kinematic equation of position as a function of time.

Solution: The position kinematic equation is $x=\frac 12\,a\,t^{2}+v_0\,t+x_0$. First, find the acceleration as below \begin{align*} v^{2}-v_0^{2}&=2\,a\,\Delta x\\8^{2}-4^{2}&=2\,a\,(10-4)\\\Rightarrow a&=4\,{\rm m/s^{2}}\end{align*}Now plug the known values in the position equation \begin{align*}x&=\frac 12\,a\,t^{2}+v_0\,t+x_0\\&=\frac 12\,(4)t^{2}+4\,t+4\\&=2t^{2}+4\,t+4\end{align*}

Problem (35): Velocity of an object as a function of time is as $v=2\,t+4$. What is its total displacement after $2\,{\rm s}$?

Solution: By comparing those with the velocity kinematic equation $v=v_0+a\,t$, one can identify acceleration and initial velocity as $4\,{\rm m/s}$,$2\,{\rm m/s^{2}}$,respectively. Now applying displacement kinematic formula $\Delta x=\frac 12\,a\,t^{2}+v_0\,t$ at time $t_2=2\,{\rm s}$ to find the total displacement \begin{align*}\Delta x&=\frac 12\,a\,t^{2}+v_0\,t+x_0\\\Delta x&=\frac 12\,(2)\,(2)^{2}+4(4)\\&=20\,{\rm m}\end{align*}

Problem (36): A bus in a straight path accelerates and travels the distance of $80\,{\rm m}$ between $A$ and $B$ in $8\,{\rm s}$. At $B$, its speed becomes $15\,{\rm m/s}$. What is the acceleration of the bus?

Solution:
known values: displacement $\Delta x_{AB}=80\,{\rm m}$, $\Delta t=8\,{\rm s}$, $v_B=15\,{\rm m/s}$, acceleration $a=?$
Once the initial velocity given the displacement is obtained by $\Delta x=\frac 12\,at^{2}+v_0\,t$ and once the final velocity is given the displacement gets by kinematic equation $\Delta x=-\frac 12\,at^{2}+v_f\,t$. In this problem, the velocity at the end of the path is given so we have \begin{align*}\Delta x&=-\frac 12\,at^{2}+v_f\,t\\80&=-\frac 12\,a\,(8)^{2}+(15)(8)\\\Rightarrow a&=-\frac{40}{32}\\&=-\frac 54\end{align*}

Problem (37): Starting from rest and at the same time, two objects with accelerations of $2\,{\rm m/s^2}$ and $8\,{\rm m/s^2}$ travel from $A$ in a straight line to $B$. If the arriving time difference between them is $3\,{\rm s}$, then how far is the total distance between $A$ and $B$?

Solution: Since faster object arrives sooner, so let the total time between $A$ and $B$ be $t$ and consequently the arriving time for slower object would be $t-3$. Now, write down the displacement kinematic equations $\Delta x=\frac 12\,a\,t^{2}+v_0\,t$ for two objects and equate them (since their total displacement are the same)\begin{align*}\Delta x_1&=\frac 12\,(8)(t-3)^{2}+0\\\Delta x_2&=\frac 12\,(2)t^{2}+0\\\Delta x_1&=\Delta x_2\\4(t-3)^{2}&=t^{2}\end{align*} Rearranging and simplifying the above equation we get $t^{2}-8t+12=0$. Applying the quadratic formula, $t_{1,2}=\frac{-b\pm\sqrt{b^{2}-4\,a\,c}}{2a}$ for the standard equation $at^{2}+b\,t+c=0$, we obtain two roots as $t_1=2\,{\rm s}$ and $t_2=6\,{\rm s}$. (The above roots can be obtained readily by taking square root from both sides as $t=\pm\,2(t-3)$ and solving for $t$). The accepted time is $t_2$. By plugging it into one of the displacement equations above, the total distance, which is the magnitude of the total displacement, is obtained \begin{align*}\Delta x_1&=\frac 12\,(8)(t-3)^{2}+0\\&=\frac 12\,(8)(6-3)^{2}\\&=36\,{\rm m}\end{align*}

Problem (38): The position-time equation of a moving particle is as $x=2t^{2}+3\,t$.
(a) Find its acceleration and initial velocity.
(b) What is the total distance traveled in the third second of the motion?

Solution:
(a) comparing above equation with the standard position kinematic equation $\Delta x=\frac 12\,a\,t^{2}+v_0\,t$, one can identify acceleration and initial velocity as $\frac 12\,a=2\Rightarrow a=4\,{\rm m/s^2}$ and $v_0=3\,{\rm m/s}$ ,respectively.

(b) Third second of the motion means the time interval [$t_3=3\,{\rm s},t_2=2\,{\rm s}$], so substituting these times into the equation above, the corresponding distances are given as \begin{align*}x_3&=2\,(3)^{2}+3\times 3\\&=27\,{\rm m}\\x_2&=2\,(2)^{2}+3\times 2\\&=14\,{\rm m}\\\Rightarrow \Delta x&=27-14=13\,{\rm m}\end{align*}

Problem (39): A bullet is fired from a riffle and strike a block of wood with depth of $10\,{\rm cm}$ at velocity of $400\,{\rm m/s}$ and emerges with $100\,{\rm m/s}$ from the other side of the block. Suppose that the bullet's path through the block is straight line.
(a) Find the acceleration of the bullet in the block.
(b) How long does it take the bullet passes through the block?

Solution:
(a) Consider the entry and exit velocities as the initial and final velocities, respectively. The distance between these points is also $\Delta x=10\,{\rm cm}=0.1\,{\rm m}$, so use the time-independent kinematic equation below to find the desired acceleration \begin{align*} v^{2}-v_0^{2}&=2\,a\,\Delta x\\(100)^{2}-(400)^{2}&=2\,a\,(0.1)\\\Rightarrow a&=\frac{10^{4}-16\times 10^{4}}{0.2}\\&=-7500\,{\rm m/s^2}\end{align*}

(b) With above known values, it is better to use the equation $\Delta x=\frac{v_1+v_2}2\,\Delta t$ to find the time needed as\begin{align*}\Delta x&=\frac{v_1+v_2}2\,\Delta t\\0.1&=\frac{100+400}2\,\Delta t\\\Rightarrow \Delta t&=4\times 10^{-4}\,{\rm s}\end{align*}

Problem (40): A car moving at a velocity of $72\,{\rm km/h}$ suddenly brakes and with a constant acceleration $4\,{\rm m/s^2}$ travels some distance until coming to a complete stop. Determine the time and distance traveled between braking and stopping points.

Solution: at the moment of braking, the earlier constant velocity serves as initial velocity (which must be converted into SI units $m/s$). Write the velocity kinematic equation $v=v_i+a\,t$ and substitute the known values above into it to find the time required as \begin{align*}v&=v_i+a\,t\\0&=20+(-4)\,t\\\Rightarrow a&=5\,{\rm m/s^2}\end{align*}where in the above we converted $km/h$ to $m/s$ by multiplying it by $\frac{10}{36}$. Since the velocity of the car is decreasing, so its acceleration must be negative $a=-4\,{\rm m/s^2}$.

The distance traveled is also obtained using time-independent kinematic equation $v^{2}-v_i^{2}=2\,a\,\Delta x$ as \begin{align*}v^{2}-v_i^{2}&=2\,a\,\Delta x\\0-(20)^{2}&=2(-4)\Delta x\\\Rightarrow \Delta x&=50\,{\rm m}\end{align*}

Problem (41): A plane starts moving along a straight-line path from rest and after $45\,{\rm s}$ takes off with a velocity $80\,{\rm m/s}$. Suppose the acceleration is constant across the path. Determine
(a) Plane's acceleration.
(b) the distance which the plane travels before taking off the ground.

Solution:
(a) Kinematic velocity equation $v=v_0+a\,t$ gives the unknown acceleration \begin{align*}v&=v_0+a\,t\\80&=0+a\,(45)\\\Rightarrow a&=\frac {16}9\,{\rm m/s^{2}}\end{align*}

(b) Kinematic position equation $\Delta x=\frac 12\,a\,t^{2}+v_0\,t$ gives the magnitude of the displacement as distance traveled \begin{align*}\Delta x&=\frac 12\,a\,t^{2}+v_0\,t\\\Delta x&=\frac 12\,(16/9)(45)^{2}+0\\&=1800\,{\rm m}\end{align*}

Problem (42): An object is moving with constant speed along a straight-line path. If the object at $t_1=5\,{\rm s}$ is at position $x_1=+6\,{\rm m}$ and at $t_2=20\,{\rm s}$ is at $x_2=36\,{\rm m}$ then find its equation of position as a function of time.

Solution: Kinematic equation of position with constant speed is as $x=x_0+vt$, where $x_0$ is the initial position at time $t=0$ where the moving particle starts its motion. Substituting the time $t=5\,{\rm s}$ and position $x=+6\,{\rm m}$ into it gives $6=x_0+5v$, at time and position $t=20\,{\rm s}$ and $x=+36\,{\rm m}$ we get $36=x_0+20v$. Equating these equations results in a system of two equations with two unknowns as below $\left\{\begin{array}{rcl} 6&=&5v+x_0\\36 & = & 20v+x_0 \end{array}\right.$ Solving for unknowns, we get $v=2\,{\rm m/s}$ and $x_0=-4\,{\rm m}$. Therefore, the position verses time is as $x=2t-4$.

Problem (43): An object is moving along $x$-axis. At $t_0=0$, it passes the position $x=+4\,{\rm m}$ with a velocity of $+3\,{\rm m/s}$. If the object at $t=4\,{\rm s}$ is at the greatest distance from the origin, then at the instant of $t=8\,{\rm s}$ it is at what distance of origin?

Solution: The greatest distance from origin without changing direction means that the objects at this moment stops and changes its direction. Thus, substitute the known values $v_0=3\,{\rm m/s}$ and $v=0$ at time $t=4\,{\rm s}$ into the velocity kinematic equation $v=v_0+at$ to find the acceleration of the object. \begin{align*}v&=v_0+at\\0&=3+a\,(4)\\\Rightarrow a&= -\frac 34\,{\rm m/s^2}\end{align*} Now write down the position kinematic equation $x=\frac 12\,at^{2}+v_0t+x_0$ to find the position as a function of time as \begin{align*}x&=\frac 12\,at^{2}+v_0t+x_0\\&=\frac 12\,(-\frac 34)t^{2}+3t+4\\&=-\frac 38\,t^{2}+3t+4\end{align*} Now at time $t=8\,{\rm s}$ its position is \begin{align*}x&=-\frac 38\,t^{2}+3t+4\\&=-\frac 38\,(8)^{2}+3(8)+4\\&=4\,{\rm m}\end{align*}

Problem (44): A stone is thrown vertically upward from a building of $15\,{\rm m}$ high with an initial velocity of $10\,{\rm m/s}$. What is the stone's velocity just before hitting the ground?

Solution: In the free-fall problems, the important note is choosing the origin. Usually, the throwing (releasing or dropping) point is the best choice. In this case, the hitting point is below the origin so its vertical displacement ($y$) is negative.

Applying the time-independent free fall kinematic equation, we have \begin{align*}v_f^{2}-v_i^{2}&=2\,(-g)\Delta y\\v_f^{2}-(10)^{2}&=2\,(-10)(-15)\\\Rightarrow v_f&=\pm 20\,{\rm m/s}\end{align*} Since velocity is a vector quantity and just before striking to the ground its direction is vertically downward so the negative value must be chosen i.e. $v_f=-20\,{\rm m/s}$.

Practice Problem (45): A bullet is released without initial velocity from a height of $h$ and in the last second of falling, it drops a distance of $35\,{\rm m}$. Determine the height of $h$?

Solution: this left for you as a practice problem.

Problem (46): A bullet is fired with an initial velocity of $15\,{\rm m/s}$ from the top of a tower of $20\,{\rm m}$ high. What is its velocity at the instant of striking at the ground?

Solution: Let the origin be the firing point. Using the below kinematic equation we have \begin{align*}v_f^{2}-v_i^{2}&=2(-g)\Delta y\\v_f^{2}-0&=2(-10)(-20)\\\Rightarrow v_f&=-20\,{\rm m/s}\end{align*} where we chose the minus sign as the direction of velocity is downward.

Problem (47): There is a well with a depth of $34\,{\rm m}$. A person drops a stone vertically into it with an initial velocity of $7\,{\rm m/s}$. What is the time interval between dropping the stone and hearing its impact sound? (Assume $g=10\,{\rm m/s^2}$ and the speed of the sound in the air is $340\,{\rm m/s}$).

Solution: This motion problem has two parts. One is descending into the well which is a constant acceleration motion and the other is ascending the sound of impact which is uniform motion.

For the first part use the kinematic equation $\Delta y=-\frac 12 gt^{2}+v_0 t$ to find the falling time as \begin{align*}\Delta y&=-\frac 12 gt^{2}+v_0 t\\-34&=-\frac 12 (10)t^{2}+(-7)t\\\Rightarrow 0&=5t^{2}+7t-34\end{align*}Since initial velocity is a vector and its direction is initially downward so a negative is included in front of it. The above quadratic equation has two solutions as $t_1=2\,{\rm s}$ and $t_2=-3.4\,{\rm s}$. The negative value is rejected since the initial time is $t_0=0$.

The second part is a uniform motion since the speed of sound is constant so we can calculate its rising time as \begin{align*} t&=\frac {\Delta y}{v}\\&=\frac{34}{340}=0.1\,{\rm s}\end{align*} Thus, the total time is obtained as $t_T=2+0.1=2.1\,{\rm s}$.

Problem (48): A stone is dropped vertically from the top of a building with a height of $h$. What is its speed at the height of $\frac h2$?

Solution: Let the dropping point be the origin so in the kinematic equations the vertical displacement must be negative i.e. $\Delta y=-\frac h2$. Thus, use the below equation to find the speed at the desired level \begin{align*}v_2^{2}-v_1^{2}&=2(-g)\Delta y\\v_2^{2}-0&=-2g(-\frac h2)\\\Rightarrow v_2&=\sqrt{gh}\end{align*}

Practice Problem (49): An small bullet is released without initial velocity from a tower and travels the last $80\,{\rm m}$ of its motion in $2\,{\rm s}$. What is the height of the tower?

Solution: this left for you as a practice problem.

Problem (50): A bullet is fired vertically upward from a height of $90\,{\rm m}$ from the ground surface and after $10\,{\rm s}$ reaches the ground. After $2\,{\rm s}$ from the throwing point, the bullet is how far away from the surface? ($g=9.8\,{\rm m/s^2}$)

Solution: In a free-fall problem, the vertical position of an object in an instant of time is given by kinematic equation $y=-\frac 12 gt^{2}+v_0 t+y_0$. Let the firing point be origin so $y_0=0$. To complete the equation above, you must have an initial speed.

To find the initial speed, apply kinematic formula $y=-\frac 12 gt^{2}+v_0 t+y_0$ between origin and striking point.\begin{align*}y&=-\frac 12 gt^{2}+v_0 t+y_0\\-90&=-\frac 12 \,(9.8)(10)^{2}+v_0\,(10)+0\\\Rightarrow v_0&=40\,{\rm m/s}\end{align*} The striking point is $-90\,{\rm m}$ below the origin.

Given initial velocity, substitute it into the above equation again with time $t=2\,{\rm s}$ to find the position of the bullet after $2\,{\rm s}$ with respect to the firing point \begin{align*}y&=-\frac 12 gt^{2}+v_0 t+y_0\\&=-\frac 12\,(9.8)(2)^{2}+(40)(2)+0\\\Rightarrow y&=60.4\,{\rm m}\end{align*}

Problem (51): A ball is thrown vertically upward with an initial velocity of $18\,{\rm m/s}$. How many seconds after throwing, the ball's speed is $9\,{\rm m/s}$ downward?

Solution: In all kinematic equations, $x,y,v,v_0$ and $a$ are vectors so its signs matters. Speed of $9\,{\rm m/s}$ downward means velocity of $-9\,{\rm m/s}$. Downward or upward indicate the direction of velocity. Now use the equation $v_f=v_i-g\,t$ to find the velocity in any later time.\begin{align*}v_f&=v_i-g\,t\\-9&=+18-(10)\,t\\\Rightarrow t&=2.7\,{\rm s}\end{align*}

Practice Problem (52): A ball is released from a height. In the first second, it covers a distance of $\Delta x_1$, and in the second ones the distance $\Delta x_2$. Find the ratio of $\frac {\Delta x_2}{\Delta x_1}$.

Solution: this left for you as a practice problem.

Problem (53): An object is thrown vertically in the air from a $100\,{\rm m}$ height with an initial velocity of $v_0$. After $5\,{\rm s}$, it reaches the ground. Determine the magnitude and direction of the initial velocity.

Solution: Let the origin be the throwing point and upward direction positive. Substitute the given total time into the vertical displacement kinematic equation $\Delta y=-\frac 12 \,t^{2}+v_0\,t$ with $\Delta y=-100\,{\rm m}$ (since the impact point is below the origin). \begin{align*} \Delta y&=-\frac 12 \,t^{2}+v_0\,t\\-100&=-\frac 12\,(10)(5)^{2}+v_0\,t\\\Rightarrow v_0&=-5\,{\rm m/s}\end{align*} the minus sign indicates initial velocity is downward with the magnitude (speed) of $5\,{\rm m/s}$.

Problem (54): From a height of $15\,{\rm m}$, a ball is kicked vertically up into the air with an initial velocity of $v_0$. It reaches the highest point of its path with an elevation of $20\,{\rm m}$ from the surface. Find the initial velocity $v_0$?

Solution: The highest point is $5\,{\rm m}$ above the kicking point. Apply the time-independent kinematic equation below to find the initial velocity \begin{align*}v_f^{2}-v_i^{2}&=2(-g)\Delta y\\0-v_i^{2}&=-2(10)(5)\\\Rightarrow v_i&=\pm 10\,{\rm m/s}\end{align*}Since ball kicked upward so we must choose the plus sign i.e. $v_i=+10\,{\rm m/s}$. In above, we used this fact that in a free fall problem, at the highest point (apex) the velocity is zero ($v_f=0$).

Recall that the projectiles are a special type of free-fall motion with a launch angle of $\theta=90$ with its own formulas.

Problem (55): From the bottom of a $25\,{\rm m}$ well, a stone is thrown vertically upward with an initial velocity $30\,{\rm m/s}$.
(a) How far is the ball out of the well?
(b) The stone before returning into the well, how many seconds was outside of the well?

Solution:
(a) Let the bottom of the well be the origin. First, we find how much distance the ball rises. Recall that the highest point is where $v_f=0$ so we have\begin{align*}v_f^{2}-v_0^{2}&=-2g\Delta y\\0-(30)^{2}&=-2(10)(\Delta y)\\&=45\,{\rm m}\end{align*} Of this height $25\,{\rm m}$ is for well's height so the stone is $20\,{\rm m}$ outside of the well.

(b) Here, we have $\Delta y=0$ and $v_0$ must be determined as below\begin{align*}v_i^{2}-v_0^{2}&=-2g\Delta y\\v_i^{2}-(30)^{2}&=-2(10)(25)\\\Rightarrow v_i&=+20\,{\rm m/s}\end{align*} where $v_i$ is the velocity just before leaving the well which can serve as initial velocity for the second part to find the total time which the stone is out of the well\begin{align*}\Delta y&=-\frac 12 gt^{2}+v_0 t\\0&=-\frac 12 (-10)t^{2}+20\,(2)\end{align*} Solving for $t$, one can obtain the required time is $t=4\,{\rm s}$.

Problem (56): From the top of a $20-{\rm m}$ tower, a small ball is thrown vertically upward. If $4\,{\rm s}$ after throwing it hit the ground, how many seconds before striking to the surface the ball meet the initial launching point again? (Air resistance is neglected and $g=10\,{\rm m/s^2}$).

Solution: Let the origin be the throwing point. The tower's height is $20-{\rm m}$ and total time which the ball is in the air is $4\,{\rm s}$. With these known values, one can find the initial velocity as \begin{align*}\Delta y&=-\frac 12 gt^{2}+v_0\,t\\-25&=-\frac 12 (10)(4)^{2}+v_0\,(4)\\\Rightarrow v_0&=15\,{\rm m/s}\end{align*} When the ball returns to its initial point, its total displacement is zero i.e. $\Delta y=0$ so we can use the following kinematic equation to find the total time to return to the starting point \begin{align*}\Delta y&=-\frac 12 gt^{2}+v_0\,t\\0&=-\frac 12\,(10)t^{2}+(15)\,t\end{align*} Rearranging and solving for $t$, we get $t=3\,{\rm s}$.

Problem (57): A rock is thrown vertically upward into the air. It reaches the height of $40\,{\rm m}$ from the surface at times $t_1=2\,{\rm s}$ and $t_2$. Find $t_2$ and determine the greatest height reached by the rock (neglect air resistance and let $g=10\,{\rm m/s^2}$).

Solution: Let the trowing point (surface of ground) be the origin. Between origin and the point with known values $h=4\,{\rm m}$, $t=2\,{\rm s}$ one can write down the kinematic equation $\Delta y=-\frac 12 gt^{2}+v_0\,t$ to find the initial velocity as\begin{align*} \Delta y&=-\frac 12 gt^{2}+v_0\,t\\40&=-\frac 12\,(10)(2)^{2}+v_0\,(2)\\\Rightarrow v_0&=30\,{\rm m/s}\end{align*}Now we are going to find the times when the rock reaches the height $40\,{\rm m}$ (Recall that when an object is thrown upward, it passes through every point twice). Applying the same equation above, we get \begin{align*}\Delta y&=-\frac 12 gt^{2}+v_0\,t\\40&=-\frac 12\,(10)t^{2}+30\,t\end{align*} Rearranging and solving for $t$ using quadratic formula, two times are obtained i.e. $t_1=2\,{\rm s}$ and $t_2=4\,{\rm s}$. The greatest height is where the vertical velocity becomes zero so we have \begin{align*}v_f^{2}-v_i^{2}&=2(-g)\Delta y\\0-(30)^{2}&=2(-10)\Delta y\\\Rightarrow \Delta y&=45\,{\rm m}\end{align*} Thus, the highest point located $H=45\,{\rm m}$ above the ground.

Problem (58): A ball is launched with an initial velocity of $30\,{\rm m/s}$ vertically upward. How long will it take to reaches $20\,{\rm m}$ below the highest point for the first time? (neglect air resistance and assume $g=10\,{\rm m/s^2}$).

Solution: Between the origin (surface level) and the highest point ($v=0$) apply the time-independent kinematic equation below to find the greatest height $H$ where the ball reaches.\begin{align*}v^{2}-v_0^{2}&=-2\,g\,\Delta y\\0-(30)^{2}&=-2(10)H\\\Rightarrow H&=45\,{\rm m}\end{align*} The point $20\,{\rm m}$ below $H$ has height of $h=45-20=25\,{\rm m}$. The time needed for reaching that point is obtained as\begin{align*}\Delta y&=-\frac 12\,g\,t^{2}+v_0\,t\\25&=-\frac 12\,(10)\,t^{2}+30\,(t)\end{align*} Solving for $t$ (using quadratic formula), we get $t_1=1\,{\rm s}$ and $t_2=5\,{\rm s}$ one for up way and the second for down way.

Practice Problem (59): A rock is thrown vertically upward from a height of $60\,{\rm m}$ with an initial speed of $20\,{\rm m/s}$. Find the ratio of displacement in the third second to the displacement in the last second of the motion?

Solution: this left for you as a practice problem.

Problem (60): Ball $A$ is released directly downward from the height of $h$. After $3\,{\rm s}$, ball $B$ is released from the height of $\frac h4$. What is the ratio of the velocity of the ball $A$ to the ball $B$ at the instant of hitting the ground?

Solution: According to the time-independent kinematic equation $v^{2}-v_0^{2}=-2\,g\,\Delta y$, the velocity at the end of the path is dependent on the height from which the object drops. Thus,\begin{align*}\frac{v_A^{2}}{v_B^{2}}&=\frac{-2\,g\,h_A}{-2\,g\,h_B}\\&=\frac{h}{h/4}\\&=4\\\Rightarrow \frac{v_A}{v_B}&=2\end{align*}

Practice Problem (61): From a $45-{\rm m}$ cliff, two balls directly downward, one is released, and the other is dropped with an initial velocity of $12.5\,{\rm m/s}$. What is the difference between their distances traveled in the last second of their motions?

Solution: this left for you as a practice problem. You can find its solution later.

Problem (62): A stone is launched directly upward from the surface level with an initial velocity of $20\,{\rm m/s}$. How many seconds after launch is the stone's velocity $5\,{\rm m/s}$ downward?

Solution:
Let the origin be the surface level and positive direction up. Therefore, we have initial velocity $v_0=+20\,{\rm m/s}$ and final velocity $v=-5\,{\rm m/s}$. Use the velocity kinematic equation $v=v_0-g\,t$ to find the time required as \begin{align*}v&=v_0-g\,t\\-5&=+20-10\times t\\\Rightarrow t&=2.5\,{\rm s}\end{align*}

Problem (63): A rock is dropped from a tower with the height $60\,{\rm m}$. At the instant that the rock reaches $15\,{\rm m}$ above the ground level, its velocity is what fraction of its velocity at the moment of hitting the ground?

Solution: Let the dropping point be the origin and positive direction up. Velocity $v_1$ at the height of $h=60-15=45\,{\rm m}$ below the origin obtained (must be included with a negative in the kinematic equation) as \begin{align*}v_1^{2}-v_0^{2}&=-2\,g\Delta y\\v^{2}-0&=-2(10)(-45)\\\Rightarrow v&=-30\,{\rm m/s}\end{align*}In above, we chose the minus sign since the rock's direction is downward. Now use again the above equation to find the velocity at the hitting point $v_2$ as \begin{align*}v_2^{2}-v_0^{2}&=-2\,g\Delta y\\v^{2}-0&=-2(10)(-60)\\\Rightarrow v&=\sqrt{1200}\\&=\sqrt{400\times 3}=20\sqrt{3}\end{align*} The asked ratio is $\frac{v_1}{v_2}=\frac{3}{2\sqrt{3}}$.

Problem (64): From a $25-{\rm m}$ building, a ball is thrown vertically upward at an initial velocity of $20\,{\rm m/s}$. How long will it take to hit the ground?

Solution: Origin is considered to be at throwing point. Applying the position kinematic equation below to find the required time \begin{align*}\Delta y&=-\frac 12\,gt^{2}+v_0\,t\\-25&=-\frac 12\,(10)\,t^{2}+20\,t\end{align*} Rearranging and converting it into the standard form of quadratic equation $ax^{2}+b\,x+c=0$ as $t^{2}-4t-5=0$, solutions are obtained as\begin{align*}x_{1,2}&=\frac{-b\pm \sqrt{b^{2}-4\,a\,c}}{2a}\\&=\frac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-5)}}{2(1)}\\&=-1 \ \text{and}\ 5 \end{align*}Therefore, the time needed the ball hit the ground is $5\,{\rm s}$.

Problem (65): From the top of a building with a height of $60\,{\rm m}$, a rock is thrown directly upward at an initial velocity of $20\,{\rm m/s}$. What is the rock's velocity at the instant of hit to the ground?

Solution: Apply the time-independent kinematic equation as \begin{align*}v^{2}-v_0^{2}&=-2\,g\,\Delta y\\v^{2}-(20)^{2}&=-2(10)(-60)\\v^{2}&=1600\\\Rightarrow v&=40\,{\rm m/s}\end{align*}Therefore, the rock's velocity when it hit the ground is $v=-40\,{\rm m/s}$.

All these kinematic problems on speed, velocity, and acceleration are easily solved by choosing an appropriate kinematic equation. Keep in mind that these motion problems in one dimension are of the uniform or constant acceleration type. Projectiles are also another type of motions in two dimensions with constant acceleration.

Publisher: PhysExams.com