Question

A car starts from rest and again comes to rest after travelling $200\,{\text{m}}$ in a straight line. If its acceleration and deceleration are limited to $10\,{\text{m/}}{{\text{s}}^2}$ and $20\,{\text{m/}}{{\text{s}}^2}$ respectively minimum time the car will take to travel the distance is:
A. $20\,{\text{s}}$
B. $10\,{\text{s}}$
C. $2\sqrt {15} \,{\text{s}}$
D. $\dfrac{{20}}{3}\,{\text{s}}$

Answer 1

Use the kinematic equation for displacement and final velocity of the object. Determine the final velocity of the car during its acceleration. Then determine the relation between the times for the acceleration and deceleration of the car. Hence, use the formula total displacement of the car and determine the times required for acceleration and deceleration.

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 the 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 the time.

Complete step by step answer:
We have given that a car starts from rest and after travelling the distance of $200\,{\text{m}}$, it comes to rest.
$s = 200\,{\text{m}}$
The acceleration and deceleration of the car are $10\,{\text{m/}}{{\text{s}}^2}$ and $20\,{\text{m/}}{{\text{s}}^2}$ respectively.
${a_1} = 10\,{\text{m/}}{{\text{s}}^2}$
${a_2} = 20\,{\text{m/}}{{\text{s}}^2}$
The initial velocity ${u_1}$ of the car during its acceleration is zero as it starts from rest.
${u_1} = 0\,{\text{m/s}}$

Let us determine the final velocity of the car during its acceleration.
Rewrite equation (2) for the final velocity of the car in its acceleration.
${v_1} = {u_1} + {a_1}{t_1}$
Here, ${t_1}$ is the time of travel for the car in its acceleration.
Substitute $0\,{\text{m/s}}$ for ${u_1}$ and $10\,{\text{m/}}{{\text{s}}^2}$ for ${a_1}$ in the above equation.
${v_1} = \left( {0\,{\text{m/s}}} \right) + \left( {10\,{\text{m/}}{{\text{s}}^2}} \right){t_1}$
$\Rightarrow {v_1} = 10{t_1}$
Rewrite equation (2) for final velocity of car in its deceleration.
${v_2} = {u_2} + {a_2}{t_2}$
Here, ${t_2}$ is the time of travel for the car in its deceleration.

The final velocity of the car is zero as it comes to rest and the initial velocity of the car in deceleration is the final velocity of the car in acceleration.
${v_2} = 0\,{\text{m/s}}$
${u_2} = 10{t_1}$
Substitute $0\,{\text{m/s}}$ for ${v_2}$, $10{t_1}$ for ${u_2}$ and $- 20\,{\text{m/}}{{\text{s}}^2}$ for ${a_2}$ in the above equation.
$\left( {0\,{\text{m/s}}} \right) = 10{t_1} + \left( { - 20\,{\text{m/}}{{\text{s}}^2}} \right){t_2}$
$\Rightarrow 10{t_1} = 20{t_2}$
$\Rightarrow {t_1} = 2{t_2}$ …… (3)
Rewrite equation (1) for the total displacement of the car during its travel.
$s = {u_1}\left( {{t_1} + {t_2}} \right) + \dfrac{1}{2}\left( {{a_1}t_1^2 + {a_2}t_2^2} \right)$

Substitute $200\,{\text{m}}$ for $s$, $0\,{\text{m/s}}$ for ${u_1}$, $2{t_2}$ for ${t_1}$, $10\,{\text{m/}}{{\text{s}}^2}$ for ${a_1}$ and $20\,{\text{m/}}{{\text{s}}^2}$ for ${a_2}$ in the above equation.
$200\,{\text{m}} = \left( {0\,{\text{m/s}}} \right)\left( {2{t_2} + {t_2}} \right) + \dfrac{1}{2}\left[ {\left( {10\,{\text{m/}}{{\text{s}}^2}} \right){{\left( {2{t_2}} \right)}^2} + \left( {20\,{\text{m/}}{{\text{s}}^2}t_2^2} \right)} \right]$
$\Rightarrow 200\,{\text{m}} = {\text{30}}{t_2}$
$\Rightarrow {t_2} = \sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$
Substitute $\sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$or ${t_2}$ in equation (3).
$\Rightarrow {t_1} = 2\sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$

The total time required for travel is
$t = {t_1} + {t_2}$
Substitute $2\sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$ for ${t_1}$ and $\sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$ for ${t_2}$ in the above equation.
$t = 2\sqrt {\dfrac{{200}}{{30}}} \,{\text{s}} + \sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$
$\Rightarrow t = 3\sqrt {\dfrac{{200}}{{30}}} \,{\text{s}}$
$\Rightarrow t = \sqrt {\dfrac{{1800}}{{30}}} \,{\text{s}}$
$\therefore t = 2\sqrt {15} \,{\text{s}}$
Therefore, the minimum time required for the travel is $2\sqrt {15} \,{\text{s}}$.

Hence, the correct option is C.

Note: One can also solve the same question by another way. One can write total displacement as the sum of displacement in acceleration and displacement in deceleration. Substitute the values of initial velocities, accelerations and times in this equation and determine the total time required for the travel as the sum of the two times in acceleration and deceleration.