Question

A bus starts from rest with a constant acceleration of $5m{{s}^{-2}}$. At the same time, a car travelling with constant velocity $50m{{s}^{-1}}$ overtakes and passes the bus. How fast is the bus travelling when they are side by side?

Answer 1

Hint: The car has initially overtaken the bus, but since the bus is accelerating and the car is moving at a constant pace, the separation between them decreases. There comes a point when the bus has finally overtaken the car in some time. We will first calculate this time. And then, use this time to calculate the speed of the bus when it is side by side of the car.

Complete step-by-step answer:
Let us first assign some terms that we are going to use in our calculations.
Let the constant speed of the car be denoted by ‘v’. Then its value is given to us as:
$\Rightarrow v=50m{{s}^{-1}}$
Since the initial velocity of the bus is zero, we will move on to assign the acceleration. Let the acceleration of the bus be given by ‘a’.
Now, we can say that, when the bus and the car are side by side again, they must have travelled the same distance as they are moving in a straight-line path. Let the time taken to cover this distance be ‘t’. Then, we can write:
$\begin{align}
  & \Rightarrow 50t=0+\dfrac{1}{2}\times 5\times {{t}^{2}} \\
 & \therefore t=20s \\
\end{align}$
Now, let the speed of the bus at this instant when they are side by side be ${{v}_{f}}$
Thus, the speed of the bus after 20 seconds can be calculated as:
$\begin{align}
  & \Rightarrow {{v}_{f}}=0+5\times 20 \\
 & \therefore {{v}_{f}}=100m{{s}^{-1}} \\
\end{align}$
Hence, the bus is travelling at a speed of $100m{{s}^{-1}}$ when the bus and the car are side by side to each other.

Note: It is a simple problem of straight-line motion with constant acceleration. One might confuse velocity being the same for both of them at the time of crossing but if that would have been the case then they’d still have some separation left between them. So, one should analyze the situation mentioned in the question carefully then proceed ahead.