Problem : When brakes are applied to a bus the retardation produced is 25 centimetre per second square (cm/s^{2} )and the bus takes 20 seconds to stop. Calculate a)the initial velocity of bus b) the distance traveled by bus during this time

Data given:

Retardation (-a) = – 25 cm/s^{2}

Time taken to get stopped (t_{av})=20 second

As the bus starts from rest,

∴ final velocity (v)= 0

Initial velocity (u)= ?

Distance(s) = ?

First we will calculate initial velocity of the bus.

We know

a = (v – u ) / t

Or – 25 = (0 – u )/ 20

Or -25 = – u /20

Or u = 25x 20

∴ u = 500 cm/s

Or u = (500/100) m/ s

Or u= 5 m/s

Again We know that:

s = ut + 1/2 a. t^{2}

or S = 500x 20 + 1/2 .(-25). (20)^{2}

or S = 10000 – 1/2 .25. 400

or S = 10000 – 5000

or S = 5000 cm

or S = 50 m

∴ Initial velocity (u) = 5m/s and

Distance (s)=50m __Answer__