Question

A car is standing $200m$ behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration $2m{s^{ - 2}}$ and the car has an acceleration of $4m{s^{ - 2}}$. The car will catch up with the bus after a time of:
(A) $10\sqrt 2 s$
(B) $\sqrt {110} s$
(C) $\sqrt {120} s$
(D) $15s$

Answer 1

These types of questions are solved with the help of the three equations of motion. The time in which the car catches up to the bus will be the same for the car as well as the bus.

Complete step by step solution:
Displacement is defined as the shortest possible distance between any two positions. Whereas the rate of change of displacement with respect to time is known as velocity. Also, the rate of change of velocity with respect to time is known as acceleration.
If we don’t talk in instantaneous terms, velocity is known as the single differential of displacement. And, acceleration is defined as a single differential of velocity or double differential of displacement. This can also be written as,
$v = \dfrac{{dS}}{{dt}}$
$a = \dfrac{d}{{dt}}\left( {\dfrac{{dS}}{{dt}}} \right) = \dfrac{{{d^2}S}}{{d{t^2}}} = \dfrac{{dv}}{{dt}}$
Where, $S =$Displacement,
$t =$Time,
$v =$Velocity, and
$a =$Acceleration.
Sir Isaac Newton gave three equations of motion which stand true for all the motions in one dimension. These three equations are,
$v = u + at$
$S = ut + \dfrac{1}{2}a{t^2}$
${v^2} = {u^2} + 2aS$
Where, $v =$Final velocity,
$u =$Initial velocity,
$a =$Acceleration,
$t =$Time taken, and
$S =$Displacement.
Now in this particular case, the car is already $200m$ behind the bus. So let’s say the bus covers a displacement of $S$ when the car catches onto it in time $t$. Now for the car to catch onto the bus, it has to cover $\left( {S + 200} \right)m$ in time $t$. So for the bus,
$S = {u_B}t + \dfrac{1}{2}{a_B}{t^2}$
Where, ${u_B} =$Initial Velocity of Bus$= 0m{s^{ - 1}}$
${a_B} =$Acceleration of Bus$= 2m{s^{ - 2}}$
$\Rightarrow S = 0 \times t + \dfrac{1}{2} \times 2 \times {t^2}$
$\Rightarrow S = {t^2}$
$\Rightarrow t = \sqrt S$ . . . . . . . . . Equation $\left( 1 \right)$
Now for the car,
$S + 200 = {u_C}t + \dfrac{1}{2}{a_c}{t^2}$
Where, ${u_C} =$Initial Velocity of Car$= 0m{s^{ - 1}}$
${a_C} =$Acceleration of Car$= 4m{s^{ - 2}}$
$\Rightarrow S + 200 = 0 \times t + \dfrac{1}{2} \times 4 \times {t^2}$
$\Rightarrow S + 200 = 2{t^2}$
$\Rightarrow t = \sqrt {\dfrac{{S + 200}}{2}}$ . . . . . . . . Equation $\left( 2 \right)$
Equating both the equations,
$\sqrt S = \sqrt {\dfrac{{S + 200}}{2}}$
$\Rightarrow S = \dfrac{{S + 200}}{2}$
$\Rightarrow 2S = S + 200$
$\Rightarrow S = 200m$
Now putting this value in equation $\left( 1 \right)$,
$t = \sqrt {200} s$
$\therefore t = 10\sqrt 2 s$

Therefore, option A is the correct answer.

Note: These equations are true for one dimensional motion but for two or more dimensional motion several other parameters come into play. Also the parameters we use in the one dimensional motion are used differently in two dimensional motion. For example, in one dimensional motion, the vectors of displacement, velocity and acceleration have only two options, either they act along the direction of motion or against the direction of motion. But in two dimensional motion, all the three vectors can act in an infinite number of directions.