Problem : When brakes are applied to a bus the retardation produced is 25 centimetre per second square (cm/s2 )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/s2
Time taken to get stopped (tav)=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. t2
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

READ :  A body moving with constant acceleration | travels the distance 3 metre and 8 metre respectively in 1 second and 2 second | calculate the initial velocity and the acceleration of body | Sum Motion in one Dimension - Physics class 9

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