Kinematics in One Dimension Questions

Questions

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?

The given data are
\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?

First of all, collect the given data in the interval of accelerating
\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?

the given data is
\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}\\
\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)\\
\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?

The kinematic equations of free falling motions are same as the horizontal straight-line motion but with some modifications. Here, the motions is in the vertical direction (the $y$ direction) and the acceleration is always downward with the magnitude of $a_y =-g=-10\,{\rm m/s^{2}}$.
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*}

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 : 5