Question

A car accelerates from rest with a constant acceleration of 1.5m/s^2 for 14 seconds . the car then continues traveling at the constant velocity with the speed it achieved after the 14 seconds of acceleration . it travels at this constant velocity for an additional 25 seconds. how far did the car travel during this whole motion?

Answer 1

Sol- from Question we have acceleration of car=1.5 m/s2 for 14 second initial speed was =0 m/s So distance covered by this car St S = ut + fat² where u= initial speed a= acceleration t= time Since for t4 second we have u=0 a=1.5m/st Si= X1.5X14x14 =147m Velocity of car v=ltat V = 1.5x14 = 21.0 =21ms So at the end of 14 second uelocity is=21 m/s since car continue with this constant uelocity for 25 second, conexed distances in this time S2 = Velocity & time S2 = 21X25 = 525m so total distance covered by car= 8 S=S,+S2 = 672m Hence our car travel 672 m during this whole motion