Question

A car starts from rest and moves with a constant acceleration $2\,{\text{m/}}{{\text{s}}^2}$ for 30 seconds. The brakes are then applied and the car comes to rest in another 60 seconds. Distance between two points where its speed is half of the maximum speed is:
A. $225\,{\text{m}}$
B. $625\,{\text{m}}$
C. $1350\,{\text{m}}$
D. $2025\,{\text{m}}$

Answer 1

Use the three kinematic equations for motion of the object. First determine the distance travelled by the car in the first 30 seconds and then determine the final velocity of the car in the first 30 seconds and acceleration in next 60 seconds. Determine the distance travelled by the car between two points where speed of car is half of its maximum value.

Formulae used:
The kinematic equation for displacement $s$ of an object is
$s = ut + \dfrac{1}{2}a{t^2}$ …… (1)
Here, $u$ is the initial velocity, $a$ is the acceleration and $t$ is time.
The kinematic equation for final velocity $v$ of the object is
$v = u + at$ …… (2)
Here, $u$ is the initial velocity, $a$ is the acceleration and $t$ is time.
The kinematic equation for final velocity $v$ of the object is
${v^2} = {u^2} + 2as$ …… (3)
Here, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement.

Complete step by step answer:
We have given that the car starts from rest and moves with a constant acceleration of $2\,{\text{m/}}{{\text{s}}^2}$ for $30\,{\text{s}}$.
${a_1} = 2\,{\text{m/}}{{\text{s}}^2}$
Let us first determine the distance ${s_1}$ travelled by the car in first $30\,{\text{s}}$.
Here, the initial velocity ${u_1}$ of the car is zero.
${u_1} = 0\,{\text{m/s}}$
Rewrite equation (1) for distance travelled by the car in$30\,{\text{s}}$ .
${s_1} = {u_1}{t_1} + \dfrac{1}{2}{a_1}t_1^2$
Substitute $0\,{\text{m/s}}$ for ${u_1}$, $2\,{\text{m/}}{{\text{s}}^2}$ for ${a_1}$ and $30\,{\text{s}}$ for ${t_1}$ in the above equation.
${s_1} = \left( {0\,{\text{m/s}}} \right)\left( {30\,{\text{s}}} \right) + \dfrac{1}{2}\left( {2\,{\text{m/}}{{\text{s}}^2}} \right){\left( {30\,{\text{s}}} \right)^2}$
$\Rightarrow {s_1} = 900\,{\text{m}}$
Hence, the distance travelled by the car is first $30\,{\text{s}}$ is $900\,{\text{m}}$.

Let us now determine the velocity of the car at the end of 30 seconds.
Rewrite equation (2) for the final velocity of the car.
${v_1} = {u_1} + {a_1}{t_1}$
Substitute $0\,{\text{m/s}}$ for , $2\,{\text{m/}}{{\text{s}}^2}$ for ${a_1}$ and $30\,{\text{s}}$ for ${t_1}$ in the above equation.
${v_1} = \left( {0\,{\text{m/s}}} \right) + \left( {2\,{\text{m/}}{{\text{s}}^2}} \right)\left( {30\,{\text{s}}} \right)$
$\Rightarrow {v_1} = 60\,{\text{m/s}}$
Hence, the velocity of the car at the end of 30 seconds is $60\,{\text{m/s}}$.
After $30\,{\text{s}}$ the brakes are applied and the car comes at rest in $60\,{\text{s}}$. Hence, the initial velocity and final velocity becomes $60\,{\text{m/s}}$ and zero.
${u_2} = 60\,{\text{m/s}}$
$\Rightarrow {v_2} = 0\,{\text{m/s}}$
$\Rightarrow {t_2} = 60\,{\text{s}}$

Let us determine the acceleration during 60 seconds.
Rewrite equation (2) for the final velocity of the car.
${v_2} = {u_2} + {a_2}{t_2}$
Substitute $60\,{\text{m/s}}$ for ${u_2}$, $0\,{\text{m/s}}$ for ${v_2}$ and $60\,{\text{s}}$ for ${t_2}$ in the above equation.
$\left( {0\,{\text{m/s}}} \right) = \left( {60\,{\text{m/s}}} \right) + {a_2}\left( {60\,{\text{s}}} \right)$
$\Rightarrow {a_2} = - 1\,{\text{m/}}{{\text{s}}^{\text{2}}}$
Hence, the acceleration of the car in the next 60 seconds is $- 1\,{\text{m/}}{{\text{s}}^{\text{2}}}$.
The negative sign indicates that the velocity of the car is decreasing.

From the above all calculations, we can see that the velocity of the car increases up to $60\,{\text{m/s}}$ and then decreases to become zero. Hence, the maximum speed of the car is $60\,{\text{m/s}}$. Thus, half of the maximum speed is $v = \dfrac{{60\,{\text{m/s}}}}{2} = 30\,{\text{m/s}}$.
The velocity of the car is two times in its whole journey. First when the speed of the car is increasing from zero to $60\,{\text{m/s}}$ and second when the speed of the car is decreasing from $60\,{\text{m/s}}$ to zero.

Let us determine the distance travelled $s_1^1$ by the car when its final velocity is $30\,{\text{m/s}}$.
Rewrite equation (3) for this situation.
$v_1^{2} = u_1^2 + 2{a_1}s_1^1$
Substitute $30\,{\text{m/s}}$ for $v_1^1$, $0\,{\text{m/s}}$ for ${u_1}$ and $2\,{\text{m/s}}$ for ${a_1}$ in the above equation.
${\left( {30\,{\text{m/s}}} \right)^2} = {\left( {0\,{\text{m/s}}} \right)^2} + 2\left( {2\,{\text{m/s}}} \right)s_1^1$
$\Rightarrow s_1^1 = 225\,{\text{m}}$
Hence, the distance travelled by the car when its speed is half of its maximum speed is $225\,{\text{m}}$ from the initial point.

Now let us determine the distance $s_2^1$ traveled by the car after the first 30 seconds up to its velocity decreases to $30\,{\text{m/s}}$.
Rewrite equation (3) for this situation.
$v_2^{2} = u_2^2 + 2{a_2}s_2^1$
Substitute $30\,{\text{m/s}}$ for $v_2^1$, $60\,{\text{m/s}}$ for ${u_2}$ and $- 1\,{\text{m/s}}$ for ${a_2}$ in the above equation.
${\left( {30\,{\text{m/s}}} \right)^2} = {\left( {60\,{\text{m/s}}} \right)^2} + 2\left( { - 1\,{\text{m/s}}} \right)s_2^1$
$\Rightarrow 900 = 3600 - 2s_2^1$
$\Rightarrow s_2^1 = 1350\,{\text{m}}$

The distance between the two points where the speed of the car is half of its maximum speed is given by
$s = \left( {{s_1} - s_1^1} \right) + s_2^1$
Substitute $900\,{\text{m}}$ for ${s_1}$, $225\,{\text{m}}$ for $s_1^1$ and $1350\,{\text{m}}$ for $s_2^1$ in the above equation.
$\Rightarrow s = \left( {900\,{\text{m}} - 225\,{\text{m}}} \right) + \left( {1350\,{\text{m}}} \right)$
$\Rightarrow s = \left( {675\,{\text{m}}} \right) + \left( {1350\,{\text{m}}} \right)$
$\therefore s = 2025\,{\text{m}}$
Therefore, the distance between two points where the speed is half of the maximum speed of the car is $2025\,{\text{m}}$.

Hence, the correct option is D.

Note: The students may consider the distance between the two points where the speed of the car is half of its maximum value as the sum of the distances $225\,{\text{m}}$ and . But this will not give the distance between these two points. The distance is the sum of distances $675\,{\text{m}}$ and $1350\,{\text{m}}$ because the distance $225\,{\text{m}}$ is from the origin.