Question

A bus starts from rest with a constant acceleration of 5m/s .At the same time a car travelling with a constant velocity of 50m/s overtakes and passes the bus. Find at what distance will the bus overtakes the car. how fast will the bus be travelling then?

Answer 1

Here's your answer. Check the photo.

Answer 2

Answer:

The bus is travelling at 50 m/s

Given:

Initial velocity u = 0

Acceleration = 5 $m/s^2$

For car, velocity = 50 m/s

Solution:

The bus starts from rest and travels at a constant acceleration, while a car travels with a constant velocity and is being able to overtake the bus, so the bus will overtake at some time later due to its variable velocity (constant acceleration) at a distance which  is given by,

Equation of motion, $v = u + at$

So, for the bus, we have,

$v = u + a t$

$\mathrm{v}=0+5 \times \mathrm{t}$

For car, we have,

$v = 50$

From second equation of motion,

$s=u t+\frac{1}{2} a t^{2}$

For bus,  

$s=u t+\frac{1}{2} a t^{2}$

$s=(0 \times t)+\frac{1}{2}\left(5 \times t^{2}\right)$

$\mathrm{s}=\frac{5}{2 \mathrm{t}^{2}}$

For car,

$s=v \times t$

$\mathrm{s}=50 \mathrm{t}$

To overtake car by bus, their displacement be equal at some time,

$\frac{5}{2 \mathrm{t}^{2}}=50 \mathrm{t}$

$\mathrm{t}^{2}=2 \times 10 \mathrm{t}$

$t = 20 \ sec$

Therefore, the distance at which bus overtakes the car is given below:

$s=50 \times t$

$s=50 \times 20$

$s=1000 \ m$

Bus be travelling at that time with a velocity is given below:

$\mathrm{v}=\frac{\mathrm{s}}{\mathrm{t}}$

$\mathrm{V}=\frac{1000}{20}$

$v=50 \ \mathrm{m} / \mathrm{s}$