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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