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A car was moving with avelocity of 30 m/s. the driver braked and the car stopped after 2 seconds.assuming that the acceleration of the car was constant during the pressure period , find the acceleration and the distance travelled by the car between the time the driver pressed the brakes and the stopping time
initial velocity = u = 30m/s
final velocity = v= 0
accleration = a
we have , v = u + at
0 = 30 + 2a
or a = -30/2 = -15m/s2 (retardation)
now , we have
v2 = u2 + 2as
v=0 , u =30 , a=-15
0 = 302 + 2*(-15)s
s = 30m
retardation = -15m/s2 & distance covered before stopping is 30m
Dear Student
As acceleration is constant
Apply V=u+at
here final velocity V=0
a=-u/t = -30/2=-15 m/s2
-ve sign denotes deceleration (acceleration opposite to the direction of initial velocity)
now apply S=ut + at2/2
S=30*2 -15*22/2 = 60-30= 30m
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
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