A bus moving with speed of 36 km/h and the brakes were applied and it ...
Given , a bus is moving at a speed of 36km/h and it stops after covering 5 metre .
So , if 36km/s is taken into m/s , then it will be -
36×1000m÷60×60s
=10ms
Again , 72km/h is taken into m/s , then it will be
72×1000m÷60×60s
=20m/s
Since , 72km is double of 36km
& 20m is double of 10m
So , when the bus will move at the speed of 72km/h ,
then it will stop after covering (5×2)m , i.e. , 10m .
Because 10m is double of 5m(since it is the distance covered by the bus in the speed of 36km/h) .
Ans : The bus will cover 10m before stopping with the speed of 72km/h .
A bus moving with speed of 36 km/h and the brakes were applied and it ...
Calculating Stopping Distance of the Bus:
- Initial speed of the bus = 36 km/h
- Initial speed of the bus in m/s = 36 * (1000/3600) = 10 m/s
- Final speed of the bus = 0 m/s
- Distance covered before stopping = 5 meters
Using the equation of motion:
- v^2 = u^2 + 2as
- (0)^2 = (10)^2 + 2(-a)(5)
- 0 = 100 - 10a
- a = 10 m/s^2
Calculating stopping distance for the bus at 72 km/h:
- Initial speed of the bus = 72 km/h
- Initial speed of the bus in m/s = 72 * (1000/3600) = 20 m/s
- Using the same acceleration value obtained earlier (a = 10 m/s^2)
Using the equation of motion:
- v^2 = u^2 + 2as
- (0)^2 = (20)^2 + 2(-10)(s)
- 0 = 400 - 20s
- s = 20 meters
Therefore, the bus will cover a distance of 20 meters before stopping when moving at a speed of 72 km/h after the brakes are applied.
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