Question

A car accelerates from rest at a constant rate of 2ms 2 for some time. Then it starts decelerating at a
constant rate of 4 ms 2 and comes to rest. If it remains in motion for a total time of 6 seconds then,
(A) Its maximum speed is 8 ms -1
(B) It has travelled a total distance of 24 m
(C) Its maximum speed is 6 ms'
(D) It has travelled a total distance of 18 m

Answer 1

Answer:

Max speed is 8m/s

Explanation:

The car remains in motion for a total of 6 seconds, and since the deceleration is double the acceleration, time taken to decerate will be half the time it spends in acccelration, so 4 seconds accelerating, and 2 seconds decelerating

V = u + at = 0 + 2m/$s^{2}$ * 4 s = 2m/s ( initial speed is 0, starting from rest)

Distance travelled while accelerating = S = ut+1/2a$t^{2}$ = 0(4)+1/2*2*(4)^2

=16 m

V = u + at = 0 = 2 - 4 t^2

and while decelerating since deceleration is is -4m/s^2 is

S = -2(2)+1/2(4)(2)^2 = -4 + 8 = 4m

So total distance travelled by the car is 20m

Max speed will be at the time, the car starts to decelrate, i.e after 4 sec

its speed at that time will be v = u + at = 0 + 2m/$s^{2}$ * 4 s = 8m/s