Question

A driver of a bus moving with the speed of 50 km/h decides to stop the bus in 10
minutes. At what distance before the stop point should he apply the break

Answer 1

Answer:

Given,

u=50km/h=60×6050×1000=13.89m/s

v=0m/s

s=6m

From 2nd equation of motion,

2as=v2−u2

a=2sv2−u2

a=2×602−13.89×13.89

a=−16m/s2

Now, u=100km/h=60×60100×1000=27.78m/s

v=0m/s

From 2nd equation of motion,

2as=v2−u2

s=2av2−u2

s=−2×1602−27.78×27.78

s=24m

Explanation:

I hope that it will help you