40+ Solved Speed, Velocity, and Acceleration Problems
Simple problems on speed, velocity, and acceleration with descriptive answers are presented for the AP Physics 1 exam and college students. In each solution, you can find a brief tutorial.
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 defined in physics as 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 the 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 gives $t=1.11\,{\rm h}$.
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 $T=5+7+4=16$ minutes.
But keep in mind that since the distance is in SI units, so the time traveled must also be in SI units, which is $\rm s$. Therefore, we have\begin{align*}\text{average speed}&=\frac{\text{total distance} }{\text{total time} }\\ \\ &=\frac{350\,{\rm m}}{16\times 60\,{\rm s}}\\ \\&=0.36\,{\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, $\bar{v}=\frac{\Delta x}{\Delta t}$, is displacement divided by the elapsed time. Displacement is also a vector that obeys the addition vector rules. Thus, in this velocity problem, add each displacement to get the total displacement.
In the first part, displacement is $\Delta x_1=750\,\hat{j}$ and in 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*} In addition, the total elapsed time is $t=12\times 60$ seconds.Therefore, the magnitude of the average velocity is $\bar{v}=\frac{790.5}{12\times 60}=1.09\,{\rm m/s}$
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 in the same direction with $20\,{\rm m/s}$ for $3\,{\rm s}$. What is its average speed?
Solution: Average velocity is displacement divided by elapsed time, i.e., $\bar{v}\equiv \frac{\Delta x_{tot}}{\Delta t_{tot}}$.
In this velocity problem, the object goes through two stages with two different displacements, so add them to find the total displacement. Thus,\[\bar{v}=\frac{x_1 + x_2}{t_1 +t_2}\] Again, to find the displacement, we use the same equation as the average velocity formula. Thus, 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}\\ \\&=\boxed{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 at $600\,{\rm km/h}$. What is the flight time of the second plane?
Solution: first find the distance between two cities using the average velocity formula $\bar{v}=\frac{\Delta x}{\Delta t}$ as below \begin{align*} x&=vt\\&=900\times 1.5\\&=1350\,{\rm km}\end{align*} where we wrote one hour and a half minutes as $1.5\,\rm h$. Now use again the same kinematic equation above to find the time required for another plane \begin{align*} t&=\frac xv\\ \\ &=\frac{1350\,\rm km}{600\,\rm km/h}\\ \\&=2.25\,{\rm h}\end{align*} Thus, the time for the second plane is $2$ hours and $0.25$ of an hour, which converts to 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 $x_1=4\,{\rm m}$ and $x_2=-8\,{\rm m}$. Find the average velocity and 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 physics. In the 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 sign shows the direction of the velocity, which is in the same direction as the 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 during which the change 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 moves from rest to a speed of $45\,\rm m/s$ in a time interval of $15\,\rm s$. At what rate does the car accelerate?
Solution: The car initially is at rest, $v_1=0$, and finally reaches $v_2=45\,\rm m/s$ in a time interval $\Delta t=15\,\rm s$. Average acceleration is the change in velocity, $\Delta v=v_2-v_1$, divided by the elapsed time $\Delta t$, so \[\bar{a}=\frac{45-0}{15}=\boxed{3\,\rm m/s^2} \]
Problem (10): 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: Let the car's uniform velocity be $v_1$ and its final velocity $v_2=0$. Average acceleration is the 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}\\\\ &=\boxed{-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 (11): 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 is defined as the 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 $\rm km/h$ to the SI unit of velocity ($\rm m/s$) as \[1\,\frac{km}{h}=\frac {1000\,m}{3600\,s}=\frac{10}{36}\, \rm m/s\] so we get
\[72\,\rm km/h=72\times \frac{10}{36}=20\,\rm m/s\]
Problem (12): A race car accelerates from an initial velocity of $v_i=10\,{\rm m/s}$ to a final velocity of $v_f = 30\,{\rm m/s}$ in a time interval of $2\,{\rm s}$. Determine its average acceleration.
Solution: A change in the velocity of an object $\Delta v$ over a time interval $\Delta t$ is defined as an average acceleration.
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 (13): 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}\\&=\boxed{-15\,{\rm m/s^2}}\end{align*}Recall that in the definition above, velocities are vector quantities. The final velocity is in the opposite direction from the initial velocity so a negative must be included.
Problem (14): A ball is thrown vertically up into the air by a boy. After $4$ seconds, it reaches the highest point of its path. How fast does the ball leave the boy's hand?
Solution: At the highest point, the ball has zero speed, $v_2=0$. It takes the ball $4\,\rm s$ to reach that point. In this problem, our unknown is the initial speed of the ball, $v_1=?$. Here, the ball accelerates at a constant rate of $g=-9.8\,\rm m/s^2$ in the presence of gravity.
Using the average acceleration formula $\bar{a}=\frac{\Delta v}{\Delta t}$ and substituting the numerical values into this, we will have \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{0-v_1}{4} \\\\ \Rightarrow \boxed{v_1=39.2\,\rm m/s} \end{gather*} Note that $\Delta v=v_2-v_1$.
Problem (15): A child drops crumpled paper from a window. The paper hit the ground in $3\,\rm s$. What is the velocity of the crumpled paper just before it strikes the ground?
Solution: The crumpled paper is initially in the child's hand, so $v_1=0$. Let its speed just before striking be $v_2$. In this case, we have an object accelerating down in the presence of gravitational force at a constant rate of $g=-9.8\,\rm m/s^2$. Using the definition of average acceleration, we can find $v_2$ as below \begin{gather*} \bar{a}=\frac{\Delta v}{\Delta t} \\\\ -9.8=\frac{v_2-0}{3} \\\\ \Rightarrow v_2=3\times (-9.8)=\boxed{-29.4\,\rm m/s} \end{gather*} The negative shows us that the velocity must be downward, as expected!
Problem (16): A car travels along the $x$-axis for $4\,{\rm s}$ at an average velocity of $10\,{\rm m/s}$ and $2\,{\rm s}$ with an average velocity of $30\,{\rm m/s}$ 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 associated with each segment are calculated as below: \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}\\ \\ &=\boxed{20\,{\rm m/s}}\end{align*}
Problem (17): 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 velocity problem, the whole path $\Delta x$ is divided into two parts $\Delta x_1$ and $\Delta x_2$ 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 (18): 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 (19): 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 a constant velocity of $25\,{\rm m/s}$. 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*}
Read more related articles:
Kinematics Equations: Problems and Solutions
In the following section, some sample AP Physics 1 problems on acceleration are provided.
Acceleration Problems:
Problem (20): 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 (21): For $10\,{\rm s}$, the velocity of a car that 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&=\boxed{200\,{\rm m}}\end{align*}
Problem (22): A car travels along a straight line with 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&=\boxed{5\,{\rm m/s}}\end{align*}
One-Dimensional Motion Problems with Constant Acceleration
Problem (23): 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{gather*}\Delta x=\frac 12\,at^{2}+v_0 t\\ 45=\frac 12 (2)(5)^{2}+v_0 (5)\\ \Rightarrow \boxed{v_0=4\,{\rm m/s}} \end{gather*}
Problem (24): 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 a 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 (25): 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 the 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 (26): 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 (27): 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 (28): 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 coming 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, at the moment of braking, the car's velocity is known which can be chosen 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 (29): 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 velocities 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&=\boxed{-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 (30): Two cars start racing to reach the 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 the origin and to the 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 a 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 (31): 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 has 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 (32): 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 (33): 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 is left up to you as a practice problem.
Problem (34): 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 the 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 (35): An object starts moving from rest with an acceleration of $a$. After $t$ seconds, it applies 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 (36): The position-time equations of two moving objects along the $x$-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 the quadratic formula yields 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 (37): 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 (38): The 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 (39): 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 is 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 (40): 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 a faster object arrives sooner, let the total time between $A$ and $B$ be $t$; consequently, the arriving time for a 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 (41): 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 $\Delta x=\frac 12 at^{2}+v_0\,t$, 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) the 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 (42): A bullet is fired from a riffle and strikes a block of wood with a depth of $10\,{\rm cm}$ at a 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 a straight line.
(a) Find the acceleration of the bullet in the block.
(b) How long does it take the bullet to pass 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}&=2a\Delta x\\\\ (100)^{2}-(400)^{2}&=2\,a\,(0.1) \\\\ \Rightarrow a&=\frac{10^{4}-16\times 10^{4}}{0.2}\\\\ &=\boxed{-7500\,{\rm m/s^2}} \end{align*}
(b) With the 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&=\boxed{4\times 10^{-4}\,{\rm s}}\end{align*}
Problem (43): 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 the 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 car's velocity is decreasing, 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 (44): 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 that 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 (45): 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 versus time equation is as $x=2t-4$.
Problem (46): An object is moving along the $x$-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 the origin without changing direction means that the object 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 (47): 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 hitting the ground?
Solution: This is a freely falling problem. 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 motion in two dimensions with constant acceleration.
Author: Dr. Ali Nemati
Date Published: 9/6/2020
Updated: Sep 9, 2022
