Question

As a car starts to move from rest with acceleration $1.4m/{s^2}$, another car moving with a constant speed of 12m/s passes in a parallel lane. At what time span the first car can overtake the second car?
(A) 17.14s
(B) 20s
(C) 13.4s
(D) 12.4s

Answer 1

Hint: Here, we use the concept of relative motion. Instead of observing from the ground, we keep the frame of reference as any one of the cars. It means we consider that the entire problem is being observed from any one of the cars.

Complete step by step answer:
In this problem, we will use the accelerated car as the frame of reference.
Let, the accelerated car be car A and the car with uniform velocity be car B.
Now, to find relative initial velocity of A with respect to B
$\eqalign{
  & {V_{A/B}} = {V_{_A}} - {V_B} \cr
  & \Rightarrow {V_{A/B}} = 0 - 12 \cr
  & \Rightarrow {V_{A/B}} = - 12m/s \cr} $

Similarly, acceleration of A with respect to B
$\eqalign{
  & {a_{_{A/B}}} = {a_A} - {a_{_B}} \cr
  & \Rightarrow {a_{_{A/B}}} = 1.4 - 0 \cr
  & \Rightarrow {a_{_{A/B}}} = 1.4m/{s^2} \cr} $
At the point A overtakes B, the relative displacement between them should be zero.
${x_{A/B}} = {x_A} - {x_{_B}}$=0
Using equation of kinematics
${x_{A/B}} = {V_{A/B}}t + \dfrac{1}{2}a{t^2}$
$ \Rightarrow 0 = - 12t + {\dfrac{1}{2}}(1.4){t^2}$
$ \Rightarrow 12t \times 2 = 1.4t^2$
$ \therefore t = 17.14s$
Therefore, the time span the first car overtakes the second car is 17.14s.

Hence, the correct answer is option (A).

Note: Mainly, we use the concept of relative motion in which the observation is made not from ground but from one of the moving bodies. Here, we have considered the accelerating car as the frame of reference. But, we could have also considered the car with constant velocity as the frame of reference.
Secondly, we have used the concept that when the bodies have the same position, we consider the relative position to be zero and use the equation of kinematics in relative motion.