Question

when brakes are applied to a bus the retardation produced is 25 cm/s2 and the bus takes 20 s to stop. calculate the initial velocity of the bus and the distance travelled by bus during this time

Answer 1

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) = ? 
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

hope it's helpful...

Answer 2

v=u+at
0=u-25(20)
u-500=0
u=500cm/sec=500/100=5m/sec
v²-u²=2as
s=(v²-u²)/2a
s=0²-500²/-2(25)
-250000/-50=5000cm=50m