bus starts from rest with a constant acceleration of 5ms-2 At the same time a car travelling with a
constant velocity of 50 ms-1
overtake and passes the bus. (i) Find at what distance will the bus overtake
the car (ii) How fast will the bus be travelling.
Answers
Answer:
Answer:
The bus is travelling at 50 m/s
Given:
Initial velocity u = 0
Acceleration = 5 m/s^2m/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 + atv=u+at
So, for the bus, we have,
v = u + a tv=u+at
\mathrm{v}=0+5 \times \mathrm{t}v=0+5×t
For car, we have,
v = 50v=50
From second equation of motion,
s=u t+\frac{1}{2} a t^{2}s=ut+
2
1
at
2
For bus,
s=u t+\frac{1}{2} a t^{2}s=ut+
2
1
at
2
s=(0 \times t)+\frac{1}{2}\left(5 \times t^{2}\right)s=(0×t)+
2
1
(5×t
2
)
\mathrm{s}=\frac{5}{2 \mathrm{t}^{2}}s=
2t
2
5
For car,
s=v \times ts=v×t
\mathrm{s}=50 \mathrm{t}s=50t
To overtake car by bus, their displacement be equal at some time,
\frac{5}{2 \mathrm{t}^{2}}=50 \mathrm{t}
2t
2
5
=50t
\mathrm{t}^{2}=2 \times 10 \mathrm{t}t
2
=2×10t
t = 20 \ sect=20 sec
Therefore, the distance at which bus overtakes the car is given below:
s=50 \times ts=50×t
s=50 \times 20s=50×20
s=1000 \ ms=1000 m
Bus be travelling at that time with a velocity is given below:
\mathrm{v}=\frac{\mathrm{s}}{\mathrm{t}}v=
t
s
\mathrm{V}=\frac{1000}{20}V=
20
1000
v=50 \ \mathrm{m} / \mathrm{s}v=50 m/s