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A car accelerates uniformly from rest to a velocity of $101\,{\rm km/h}$ east in $8.0\,{\rm s}$. What is the magnitude of its acceleration?

\begin{eqnarray*}

\text{initial velocity} , v_0 &=& 0 \\

\text{final velocity} , v &=& 101\,{\rm Km/h} \\

&=& 101\,\frac{1000\,{\rm m}}{3600\,{\rm s}} \\

&=& 28.06\,{\rm m/s} \\

\text{time interval , t} &=& 8\,{\rm s} \\

\text{acceleration , } &=& ?

\end{eqnarray*}

The relevant kinematic equation which relates those together is $v=v_0+a\,t$. So

\begin{eqnarray*}

v &=& v_0+a\,t \\

28.06 &=& 0+a\,(8) \\

\Rightarrow a &=& \frac{28.06-0}{8} \\

&=& 3.51\,{\rm m/s^{2}}

\end{eqnarray*}

A car slows down uniformly from $30.0\,{\rm m/s}$ to rest in $7.20\,{\rm s}$. How far did it travel while decelerating?

\begin{eqnarray*}

\text{initial velocity} &=& 30\,{\rm m/s} \\

\text{final velocity} &=& 0 \\

\text{overall time} &=& 7.20\,{\rm s} \\

\text{distance} &=& ?

\end{eqnarray*}

One can solve this problem with two, direct and indirect, ways. In one way, first find the acceleration of the car and then use other kinematic equations to determine the desired quantity. So, the acceleration is obtained as

\begin{eqnarray*}

v &=& v_0+a\,t \\

0 &=& 30+a\,(7.2)\\

\Rightarrow a &\cong& -4.17\,{\rm m/s^{2}}

\end{eqnarray*}

The minus shows the slows down acceleration.

Now substitute the acceleration in one of the kinematic equations which relate those given data and have a missing value of distance, therefore

\begin{eqnarray*}

v^{2}-v_0^{2} &=& 2a(x-x_0)\\

0^{2}-(30)^2 &=& 2(-4.17)(x-0) \\

\Rightarrow x &\cong& 108\,{\rm m}

\end{eqnarray*}

where one can choose the initial position, $x_0$ as $0$.

In this kind of problems, since the acceleration is constant so we can use an special equation which is $a$ free as following

\begin{eqnarray*}

x-x_0 &=& \frac{v+v_0}{2}\,t \\

x -0 &=& \frac{0+30}{2}\,7.2 \\

&=& 108\,{\rm m}

\end{eqnarray*}

Note the subtle difference between these two ways. In the first approach, we have a approximate solution but in the second one is exact. To get a exact distance in the first solution, we must determine the car's acceleration with all decimal digits!

An object uniformly accelerates at a rate of $1.00\,{\rm m/s^{2}}$ east. While accelerating at this rate, the object is displaced $417.2\,{\rm m}$ east in $27.0\,{\rm s}$. What is the final velocity of the object?

\begin{eqnarray*}

\text{acceleration},\ a &=& 1.00\,{\rm m/s^{2}} \\

\text{displacement},\ x &=& 417.2\,{\rm m} \\

\text{overall time},\ t &=& 27\,{\rm s} \\

\text{final velocity},\ v &=& ?

\end{eqnarray*}

In all of standard kinematic equations the initial velocity $v_0$ is ubiquitous. Here, the initial velocity is not given so we can use an special equation which is $v_0$ free i.e. $x-x_0=vt-\frac{1}{2}\,a\,t^{2}$ where $v$ is the velocity at time $t$. Therefore,

\begin{eqnarray*}

x-x_0 &=& vt-\frac{1}{2}\,a\,t^{2} \\

417.2 - 0 &=& v\,(27)-\frac{1}{2}\,(1)(27)^{2} \\

\Rightarrow v &=& \frac{417.2+364.5}{27}\\

&=& 15.0\,{\rm m/s} \qquad \text{East}

\end{eqnarray*}

To find the direction of vector quantities such as displacement,velocity and acceleration, one should adopt a positive direction and then compare the sign of desired quantities with that direction. Here, we can choose the east direction as positive so the final velocity which is obtained with positive sign is toward the east.

An object accelerates uniformly from rest at a rate of $1.9\,{\rm m/s^{2}}$ west for $5.0\,{\rm s}$. Find:

(a) the displacement

(b) the final velocity

(c) the distance traveled

(d) the final speed

The given data,

\begin{eqnarray*}

\text{Initial velocity},\,v_0 &=& 0\\

\text{Acceleration},\,a &=& 1.9\,{\rm m/s^{2}} \\

\text{Time interval},\,t &=& 5\,{\rm s} \\

\end{eqnarray*}

(a) Use the following equation,

\begin{eqnarray*}

x-x_0 &=& \frac{1}{2}\,at^{2}+v_0 t \\

x- 0 &=& \frac{1}{2}\,(1.9)(5)^{2}+0(5) \\

x &=& 23.75\,{\rm m/s}

\end{eqnarray*}

In the second line, for convenience, we simply adopt the initial position ($x_0$) at time $t=0$ as $0$.

(b) the equation below gives the velocity at the end of time interval

\begin{eqnarray*}

v &=& v_0+at \\

v &=& 0+(1.9)(5)\\

&=& +9.5\,{\rm m/s} \qquad {West}

\end{eqnarray*}

(c) In a straight line motion, if the velocity and acceleration have the same signs, the speed of the moving object increases. Here, by establishing a coordinate system and choosing west as positive direction, we can see the acceleration and velocity are in the same direction. Therefore, the object moves west without any changing direction. In this type of motions, which object does not change its direction, the displacement (vector) and distance traveled is the same. Thus, as calculated in (a), the total distance is approximately $24\,{\rm m}$.

(d) As reasoning of (c), since the direction of the motion does not changes so the magnitude of vector quantities shows the value of scalar ones. Here, the magnitude of final velocity ($v=9.5\,{\rm m/s}$) is equal to the final speed.

A ball is thrown upwards with a speed of $24\,{\rm m/s}$. Take the acceleration due to gravity to be $10\,{\rm m/s^{2}}$.

(a) When is the velocity of the ball $12.0\,{\rm m/s}$ ?

(b) When is the velocity of the ball $-12.0\,{\rm m/s}$?

(c) What is the displacement of the ball at those times?

(d) What is the velocity of the ball $1.50\,{\rm s}$ after launch?

(e) What is the maximum height reached by the ball?

Now, applying the above changes to the following kinematic equation in the horizontal direction, we obtain

\begin{eqnarray*}

v_x &=& v_{0x}+ a_y t \\

v_y &=& v_{0y} + (-g)t \\

12 &=& 24 + (-10)t \\

\Rightarrow t &=& 1.2\,{\rm s}

\end{eqnarray*}

(b)

Recall that velocity is a vector, so in these equations its sign is important. Therefore,

\begin{eqnarray*}

v_y &=& v_{0y} + (-g)t \\

-12 &=& 24+(-10)t \\

\Rightarrow t &=& 3.6\,{\rm s}

\end{eqnarray*}

(c) The only equation which involves a relation between displacement and time is $y_1 -y_0 = \frac{1}{2}\,a_y t^{2}+v_{0y}t$. To solve the kinematic problems, we should first establish a coordinate system. Here, we place the origin of that coordinate system at the ground where the thrower is located. Using $v_{0y}=24\,{\rm m/s}$ and $y_0 = 0$, we have

\begin{eqnarray*}

y_1 -y_0 &=& \frac{1}{2}\,a_y t^{2}+v_{0y}t \\

y_1 -y_0 &=& \frac{1}{2}\,(-g) t^{2}+v_{0y}t \\

y_1 - 0 &=& \frac 12\, (-10)(1.2)^{2}+ (24)(1.2) \qquad \text{at time $t=1.2\,{\rm s}$} \\

\Rightarrow y_1 &=& 21.6\,{\rm m} \\

\text{And}\\

y_2 - 0 &=& \frac 12\, (-10)(3.6)^{2}+ (24)(3.6) \qquad \text{at time $t=3.6\,{\rm s}$} \\

\Rightarrow y_2 &=& 21.6\,{\rm m} \\

\end{eqnarray*}

The amount of displacement in two cases are equal! This shows that the ball is in the same height relative to the ground at times $1.2\,{\rm s}$ and $3.6\,{\rm s}$. Such a thing is possible when the object has the same velocity in different directions, as shown in the figure below.

(d) Use the following equation

\begin{eqnarray*}

v_y &=& v_{0y} + a_y t \\

v_y &=& 24+(-10)(1.5) \\

\Rightarrow v_y &=& +9\,{\rm m/s}

\end{eqnarray*}

(e) Choose the initial and final points at the beginning and the end of upward flight. First, find the time at which the ball reaches its maximum height,

\begin{eqnarray*}

v_{yf} &=& v_{0y}+a_y t \\

0 &=& 24+(-10)t_{\max} \\

\Rightarrow t_{\max} &=& 2.4\,{\rm s}

\end{eqnarray*}

$v_{yf}$ is the object's velocity at the end of climbing journey where it is zero.

Now that the maximum time is found, substitute it into the following equation to find the corresponding maximum height.

\begin{eqnarray*}

y_1 -y_0 &=& \frac{1}{2}\,a_y t^{2}+v_{0y}t \\

y_{\max} -y_0 &=& \frac{1}{2}\,(-g) t_{\max}^{2}+v_{0y}t_{\max} \\

y_{\max} -0 &=& \frac 12 (-10)(2.4)+24(2.4) \\

y_{\max} &=& 28.8\,{\rm m}

\end{eqnarray*}

A stone is thrown vertically upwards with an initial speed of $10.0\,{\rm m/s^{2}}$ from a cliff that is $50.0\,{\rm m}$ high.

(a) When does it reach the bottom of the cliff?

(b) What speed does it have just before hitting the ground?

(c) What is the total distance traveled by the stone?

Take the acceleration due to gravity to be $10\,{\rm m/s^{2}}$.

Place the origin of the coordinate system where the stone is thrown, so $y_0=0$. In kinematic problems, one should specify two points and apply the kinematic equation of motion to those.

(a) Label the bottom of the cliff as $\textcircled{c}$. Therefore, given the initial velocity and the height of cliff, one can use the following kinematic equation which relates those to the fall time.

\begin{eqnarray*}

y - y_0 &=& \frac 12\, a_y t^{2}+v_{0y} t \\

y_{\textcircled{c}}-y_0 &=& \frac 12\,(-g)t^{2}+v_{0y}t \\

(-50)- 0 &=& \frac 12\,(-10)t^{2}+10t

\end{eqnarray*}

Since the landing point is $50\,{\rm m}$ below the origin so its coordinate is $-50\,{\rm m}$. Rearranging above, we get a quadratic equation, $t^{2}-2t-10 =0$, whose solution gives the fall time.

Note : for a quadratic equation $ax^{2}+bx+c=0$, the values of $x$ which are the solution of it are given by the following relation

\[ x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \]

Therefore, using above relation we can get the fall time as

\begin{gather*}

t^{2}-2t-10 =0 \\

t=\frac{-(-2)\pm \sqrt{(-2)^{2}-4(1)(-10)}}{2(1)} \\

\Rightarrow t=4.31\,{\rm s}

\end{gather*}

(b) Substituting the fall time, computed in part (a), in the equation $v=v_0 +a_y t$ OR using the equation $v^{2}-v_0^{2}=2a_y (y-y_0)$, we can obtain the velocity at the moment of hitting to the ground.

\begin{eqnarray*}

v &=& v_0+a_y t \\

v_{\textcircled{c}} &=& v_{0y}+(-g)t \\

v_{\textcircled{c}} &=& 10+(-10)(4.31) \\

v_{\textcircled{c}} &=& -33.1\,{\rm m/s}\\

\end{eqnarray*}

OR

\begin{eqnarray*}

v^{2}-v_0^{2} &=& 2a_y (y-y_0)\\

v_{\textcircled{c}}^{2}-v_{0y}^{2} &=& 2(-g)(y_{\textcircled{c}}-y_0) \\

v_{\textcircled{c}}^{2}-(10)^{2} &=& 2(-10)(-50-0) \\

v_{\textcircled{c}}^{2} &=& -33.1\,{\rm m/s}

\end{eqnarray*}

(c) Applying the equation $v^{2}-v_{0y}^{2}=2(-g)(y-y_0)$ to find the distance traveled during climbing, then twice that value yield the total distance to the thrown point. Now add the cliff's height to find the total distance traveled by the object.

\begin{eqnarray*}

v_{\textcircled{b}}^{2}-v_{0y}^{2} &=& 2(-g)(y_{\textcircled{b}}-y_0) \\

0-(10)^2 &=& 2(-10)(y_{\textcircled{b}}-0) \\

\Rightarrow y_{\textcircled{b}} &=& 5\,{\rm m}

\end{eqnarray*}

\begin{eqnarray*}

\text{total distance traveled} &=& 2y_{\textcircled{b}}+\text{cliff's height}\\

&=& 2(5)+50\\

&=& 60\,{\rm m}

\end{eqnarray*}

where $\textcircled{b}$ is the highest point reached by the object.

A rock is thrown vertically down from the roof of $25.0\,{\rm m}$ high building with a speed of $5.0\,{\rm m/s}$.

(a) When does the rock hit the ground?

(b) With what speed does it hit the ground?

Take the acceleration due to gravity to be $10\,{\rm m/s^{2}}$.

\begin{eqnarray*}

y-y_0 &=& \frac 12\,(-g)t^{2}+v_{0y}t \\

(-25)-0 &=& \frac 12\,(-10)t^{2}+(-5)t

\end{eqnarray*}

In the end, we get a quadratic equation ,$t^{2}+t-5=0$, whose solutions give the fall time. Using the standard way of solution of quadratic equations, we have

\begin{gather*}

t^{2}+t-5=0 \\

t=\frac{-(1)\pm \sqrt{(1)^{2}-4(1)(-5)}}{2(1)} \\

\Rightarrow t=1.79\,{\rm s}

\end{gather*}

In above, for a quadratic equation $ax^{2}+bx+c=0$, we find the solutions as $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.

(b) Apply, $v=v_{0y}+(-g)t$ and substitute the time computed in (a) into it OR use, $v_y^{2}-v_{0y}^{2}=2(-g)(y-y_0)$. Therefore,

\begin{eqnarray*}

v_y &=& v_{0y}+(-g)t \\

v_y &=& (-5)+(-10)(1.79) \\

\Rightarrow v_y &=& -22.9\,{\rm m/s}

\end{eqnarray*}

OR

\begin{eqnarray*}

v_y^{2}-v_{0y}^{2} &=& 2(-g)(y-y_0) \\

v_y^{2} - (-5)^{2} &=& 2(-10)(-25-0) \\

v_y^{2} &=& 525 \\

\Rightarrow v_y &=& \pm 22.9\,{\rm m/s}

\end{eqnarray*}

Note that the square roots has two root but since the velocity vector of the rock points downward so we have to choose the negative i.e. $v_y =-22.9\, {\rm m/s}$.

A window is $1.50\,{\rm m}$ high. A stone falling from above passes the top of the window with a speed of $3.00\,{\rm m/s}$. When will it pass the bottom of the window? (Take the acceleration due to gravity to be $10\,{\rm m/s^{2}}$.)

\begin{eqnarray*}

y-y_0 &=& \frac 12\,(-g)t^{2}+v_{0y}t \\

(-1.5) - 0 &=& \frac 12\,(-10)t^{2}+(-3)t \\

-1.5 &=& -5t^{2}-3t

\end{eqnarray*}

After rearranging above equation, we arrive at $t^{2}+0.6t-0.3=0$ whose solution is obtained as

\begin{gather*}

t^{2}+0.6t-0.3=0 \\

t=\frac{-0.6 \pm \sqrt{(0.6)^{2}-4(1)(-0.3)}}{2(1)} \\

t_1 = 0.342\,{\rm s} \\

t_2 = -0.924\,{\rm s}

\end{gather*}

The above quadratic equation has two roots but the physical solution is the one with positive sign. The negative one indicates a time before we dropped the stone! Thus, we choose the positive solution i.e. $t=0.324\,{\rm s}$.

A ball is tossed with a velocity of $10\,{\rm m/s}$ directly vertically upward from the window located $20\,{\rm m}$ above the ground. Knowing that the acceleration of the ball is constant and equal to $9.81\,{\rm m/s^{2}}$ downward, determine:

(a) the velocity $v$ and elevation $y$ of the ball above the ground at any time $t$.

(b) the highest elevation reached by the ball and the corresponding value of $t$.

(c) the time when the ball will hit the ground and the corresponding velocity.

The velocity of a falling object at any later time $t$ is given by

\[ v_y = v_{0y}+(-g) t \]

Where the vertical constant acceleration $a_y$ is replaced by the always downward free-falling acceleration $-g$. Thus, substituting the numerical values in above, we get

\[ v_y = 10-9.81t \]

The displacement at that given time interval is obtained as

\[ y-y_0 = \frac 12\,(-g)t^{2}+v_{0y}t \]

putting the values, gives

\begin{eqnarray*}

y-y_0 &=& \frac 12\,(-g)t^{2}+v_{0y}t \\

y-0 &=& \frac 12\,(-9.81)t^{2}+(10)t \\

y &=& -4.905t^{2}+10t

\end{eqnarray*}

Note that the equation above gives the distance at any time relative to the throw's point.

(b) At the highest elevation, the vertical velocity of an falling object is always zero i.e. $v_y=0$. Therefore, using the above equation for the velocity at any time $t$, we have

\begin{eqnarray*}

v_y &=& v_{0y}+(-g) t \\

0 &=& +10 - 9.81t \\

\Rightarrow t &=& 1.019\,{\rm s}

\end{eqnarray*}

Now, substitute this time value into the equation of distance at any time

\begin{eqnarray*}

y-y_0 &=& \frac 12\,(-g)t^{2}+v_{0y}t \\

y-0 &=& \frac 12\,(-9.81)(1.019)^{2}+(10)(1.019) \\

y &=& 5.096\,{\rm m}

\end{eqnarray*}

Adding the height of windows, we can obtain the elevation from the ground at any time i.e. total distance $= 20+5.096=25.09\,{\rm m}$.

(c) The ball hit the ground where its coordinate is $20\,{\rm m}$ below origin that is we should set, $y=-20\,{\rm m}$ in the distance equation above and solving for the time.

\begin{eqnarray*}

y-y_0 &=& \frac 12\,(-g)t^{2}+v_{0y}t \\

(-20) - 0 &=& \frac 12(-9.81)t^{2}+10t \\

\end{eqnarray*}

Rearranging above, we get a quadratic equation, $4.905t^{2}-10t-20=0$, whose $t$ solutions are obtained as

\begin{gather*}

4.905t^{2}-10t-20=0 \\

\\

t=\frac{-(10) \pm \sqrt{(-10)^{2}-4(4.905)(-10)}}{2(4.905)} \\

\\

\Rightarrow t =

\cases{

t_1 = 3.281\,{\rm s} \cr

t_2 = -1.242\,{\rm s}

}

\end{gather*}

The negative time refers to a time before the ball is thrown! which is obviously incorrect. Thus, we choose the correct positive fall time, $t_1=3.281\,{\rm s}$.

The velocity at the moment of hitting to the ground is obtained by equations, $v_y^{2}-v_{0y}^{2}=2(-g)(y-y_0)$ or $v_y = v_{0y}+(-g)t$. Note that in the latter you should put the time fall computed previously back into it. Therefore,

\begin{eqnarray*}

v_y^{2}-v_{0y}^{2} &=& 2(-g)(y-y_0) \\

v_y^{2}-(10)^{2} &=& 2(-9.81)(-20-0) \\

v_y^{2} &=& 492.4 \\

\Rightarrow v_y &=& \pm 22.19\,{\rm m/s}

\end{eqnarray*}

The $\pm$ shows that there is two mathematical solutions which should be chosen by the physical reasoning. Since at the moment of hitting to the ground, the ball's vector velocity is downward so the correct sign is negative and thus, $v_y=-22.19\,{\rm m/s}$.

A $3.0\,{\rm Kg}$ ball is thrown vertically into the air with an initial velocity of $15\,{\rm m/s}$. The maximum height of the ball is

(a) $12\,{\rm m}$

(b) $11.5\,{\rm m}$

(c) $10.0\,{\rm m}$

(d) $9.5\,{\rm m}$

(e) $11\,{\rm m}$

\begin{eqnarray*}

v_y^{2}-v_{0y}^{2} &=& 2(-g)(y-y_0) \\

0 - (15)^{2} &=& 2(-9.8)(h_{\max}-0) \\

\Rightarrow h_{\max} &=& 11.47\,{\rm m}

\end{eqnarray*}

The correct answer is (b) which is near the above result.

An object starts from rest with an acceleration of $2.0\,{\rm m/s^{2}}$ that lasts for $3.0\,{\rm s}$. It then reduces its acceleration to $1.0\,{\rm m/s^{2}}$ that last for $5.0$ additional seconds. The velocity at the end of the $5.0\,{\rm s}$ interval is,

(a) $2\,{\rm m/s}$

(b) $3\,{\rm m/s}$

(c) $4\,{\rm m/s}$

(d) $5\,{\rm m/s}$

(e) $11\,{\rm m/s}$

\begin{eqnarray*}

v &=& v_0 + a_x t \\

v_1 &=& 0 + 2(3) \\

\Rightarrow v_1 &=& 6\,{\rm m/s}

\end{eqnarray*}

where $v_0$ and $v_1$ are the initial velocity and velocity at later time $t=2\,{\rm s}$. Now, repeat this process for the second stage

\begin{eqnarray*}

v &=& v_0 + a_x t\\

v_2 &=& v_1 + a_x t \\

v_2 &=& 6 + (1)(5) \\

&=& 11\,{\rm m/s}

\end{eqnarray*}

Thus, the object's velocity at the end of $5$ seconds is $11\,{\rm m/s}$.

An object initially traveling at a velocity of $2.0\,{\rm m/s}$ west accelerates uniformly at a rate of $1.3\,{\rm m/s^{2}}$ west. During this time of acceleration, the displacement of the object is $15\,{\rm m}$. Find:

**(a) the final velocity
(b) the final speed**

\begin{eqnarray*}

\text{Initial velocity},\,v_0 &=& 2.0\,{\rm m/s}\\

\text{Acceleration},\,a &=& 1.3\,{\rm m/s^{2}} \\

\text{Displacement},\,(x-x_0) &=& 15\,{\rm m} \\

\end{eqnarray*}

Recall that the difference between velocity and speed is in their definitions. Velocity is a vector quantity whose magnitude appears in all kinematic equations but the speed is scalar which depends on the total distance of the moving body. Since the object moving along a straight line without any changing direction, so at the end of a given time interval, its speed and velocity is the same. Therefore,

\begin{eqnarray*}

v^{2} - v_0^{2} &=& 2a_x (x-x_0) \\

v^{2} -(2)^2 &=& 2(1.3)(15) \\

v^{2} &=& 43 \\

\Rightarrow v &=& \sqrt{43}\\

&=& 6.55\,{\rm m/s}\qquad \text{West}

\end{eqnarray*}

Thus,

**(a) final velocity is $6.55\,{\rm m/s}$ toward west.
(b) final speed is $6.55\,{\rm m/s}$.**

A bungee cord is $11.0\,{\rm m}$ long. What will be the velocity of a bungee jumper just as the cord begins to stretch?

\begin{eqnarray*}

v^{2} -v_{0y}^{2} &=& 2a_y (y-y_0) \\

v^{2} -v_{0y}^{2} &=& 2(-g) (y-y_0) \\

v^{2} -0 &=& 2(-9.81)(-11) \\

\Rightarrow v &=& 14.7\,{\rm m/s}

\end{eqnarray*}

The displacement is set to be negative since we placed the origin of the coordinate system at the jumper's falling point i.e. $y_0 = 0$. Therefore, the cord's end is located $y = -11\,{\rm m}$ below the origin. The $\pm$ indicates physically the direction of velocity. Since the it is toward the falling direction, so the correct sign is minus.

How long will it take a cross-country skier traveling $5.0\,{\rm km/h}$ to cover a distance of $3.50\,{\rm km}$?

\begin{eqnarray*}

v &=& \frac{\Delta x}{\Delta t} \\

\Rightarrow t-t_0 &=& \frac{x-x_0 }{v} \\

&=& \frac{3.5\,{\rm Km}}{5\,{\rm Km/h}} \\

&=& 0.7\,{\rm h}\\

&=& 0.7 \times 3600\,{\rm s} \\

&=& 2520\,{\rm s}

\end{eqnarray*}

In the second, the definition of $\Delta$ is used and the initial values of $x_0 , t_0$ set to zero.

If a stone is thrown vertically upward with a velocity of $9.0\,{\rm, m/s}$, what is its

**(a)** Displacement after $1.5\,{\rm s}$?

**(b)** Velocity after $1.5\,{\rm s}$?

**(a)** First, adopt a coordinate system whose origin, for simplicity, is placed at the throw's point i.e. in equations, set $y_0 =0$. The given values are that of initial velocity and elapsed time and the displacement is the only unknown quantity, so the only kinematic equation which relates those together is following

\begin{eqnarray*}

y-y_0 &=& \frac 12 \,(a_y)t^{2}+v_{0y}t \\

&=& \frac 12 \,(-g)t^{2}+v_{0y}t \\

&=& \frac 12 \, (-10)(1.5)^{2}+(9)(1.5) \\

&=& 2.25\,{\rm m}

\end{eqnarray*}

**(b)** Using the equation $v_y = v_{0y}+(-g)t$, one can find the corresponding velocity at later time $t$.

\begin{eqnarray*}

v_y &=& v_{0y} +(-g)t \\

&=& 9 + (-9.81)(1.5) \\

&=& - 5.71\,{\rm m/s}

\end{eqnarray*}

The negative indicates that the direction of the stone's velocity is * downward*.

A stone is thrown vertically upward and it returns to the thrower $3.2\,{\rm s}$ later.

(a) What is the stone’s maximum displacement?

(b) What is the velocity of the stone when it is released by the thrower?

\[ t_{top} = \frac 12 \,t_{tot} \]

Here, the initial velocity, which is ubiquitous in kinematic equations, is not given so we can use the following special equation which is $v_0$-free as

\begin{eqnarray*}

y - y_0 &=& v_y t - \frac 12\, (-g)t^{2} \\

H - y_0 &=& v_{top} t_{top} - \frac 12\, (-g)t_{top}^{2} \\

H - 0 &=& 0(1.6) - \frac 12 \, (-9.81)(1.6)^{2} \\

&=& 12.55\,{\rm m}

\end{eqnarray*}

In second line, we labeled the maximum height and the corresponding velocity as $H$ and $v_{top}$. Velocity at the maximum distance is always zero i.e. $v_{top} = 0$. In addition, we placed the origin of coordinate system at throw's point so $y_0 = 0$.

(b) Using the equation $v_y = v_{0y}+(-g)t$ and substituting the known values of maximum height, $v_{top}=0$ , $t_{top}$, we can find the unknown stone's initial velocity as

\begin{eqnarray*}

v_{top} &=& v_{0y}+(-g)t_{top} \\

0 &=& v_{0y} + (-9.81)(1.6) \\

\Rightarrow v_{0y} &=& +15.7\,{\rm m/s} \qquad \text{Upward}

\end{eqnarray*}

A car is traveling north of a city street. It starts from rest at a stop light and accelerates uniformly at a rate of $1.3\,{\rm m/s^{2}}$ until it reaches the speed limit of $14\,{\rm m/s}$. The car will travel at this velocity for $3.0$ minutes and will then decelerate at a uniform rate of $1.6\,{\rm m/s^{2}}$ until it comes to stop at the next stop light. How far apart are the two lights?

\begin{eqnarray*}

\text{initial velocity}, \, v_{0I} &=& 0 \\

\text{acceleration}, \, a_I &=& 1.3\,{\rm m/s^{2}} \\

\text{final velocity}, \, v_I &=& 14\,{\rm m/s} \\

\end{eqnarray*}

With these, the only kinematic equation which relates them together and gives the unknown distance is $v^{2}-v_0^{2} = 2a\Delta x$. Therefore,

\begin{eqnarray*}

v_I^{2}-v_{0I}^{2} &=& 2a_I \Delta x_I \\

(14)^{2} - 0 &=& 2(1.3)\Delta x_I \\

\Rightarrow \Delta x_I &=& 75.38\, {\rm m}

\end{eqnarray*}

where $\Delta x_I$ is the distance traveled in first stage.

After the initial uniform acceleration motion, the car has uniform motion at constant velocity. In this stage $II$, we use the definition of average velocity to find the distance traveled as

\begin{eqnarray*}

v &=& \frac{\Delta x}{\Delta t} \\

\\

\Delta x_{II} &=& v \times \Delta t \\

&=& 14 \times (3\times 60 \,{\rm s}) \\

&=& 2520\,{\rm m}

\end{eqnarray*}

In the last stage, same as before, we have

\begin{eqnarray*}

\text{initial velocity}, \, v_{0III} &=& 14\,{\rm m/s} \\

\text{acceleration}, \, a_{III} &=& -1.6\,{\rm m/s^{2}} \\

\text{final velocity}, \, v_{III} &=& 0 \\

\end{eqnarray*}

Note that, the speed limit in the stage $II$ is used as the initial velocity of second stage and the minus shows the deceleration behavior of the motion. Thus, the distance traveled in this stage is obtained as

\begin{eqnarray*}

v_{III}^{2}-v_{0III}^{2} &=& 2a_{III} \Delta x_{III} \\

0 - (14)^{2} &=& 2(-1.6)\Delta x_{III} \\

\Rightarrow \Delta x_{III} &=& 61.25\, {\rm m}

\end{eqnarray*}

Therefore, the total distance traveled by the car between the stop lights are

\[ \Delta x_I + \Delta x_{II} + \Delta x_{III} = 2656.63\,{\rm m} \]

Category : Kinematics in One Dimension

MOST USEFUL FORMULA IN One Dimension:

Horizontal motion with constant acceleration:

\[v_x=v_0+a_xt\]

\[x=\frac 1 2 a_x t^2 +v_0t+x_0\]

\[v^2-v_0^2=2a\Delta x\]

Free falling motion:

\[v_y=v_0-gt\]

\[y=-\frac 1 2 gt^2+v_0t+y_0\]

\[v^2-v_0^2=-2g\Delta y\]

Number Of Questions : 17

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