A car starts from rest and moves with a uniform acceleration of 0.2m/s^2 for 2 minutes.then it maintains...?

Question

a unifrom velocity for 5minutes,after which the brakes are applied so as to produce a uniform retardation of 1.5m/s^2.Calculate the maximum velocity and the total distance travelled??????

Answer 1

MAximum velocity is

V = at = 0.2m/s^2 * 120s = 24 m/s

Distance traveled while accelerating
d = 1/2at^2
d = 1/2 * 0.2m/s^2 * (120s)^2 = 1440m

Distance traveled at constant speed (v=24m/s)
d = vt = 24m/s * 300s = 7200m

Distance traveled while decelerating
First let's figure out how much time to stop
0m/s = at + v
t = -v/a
t = -24m/s/(-1.5m/s^2) = 16s

Second lets compute distance
d = 1/2at^2 + vt
d = 1/2 * (-1.5m/s^2)*(16s)^2 + 24m/s * 16s
d = 192m

So total distance traveled is
1440m + 7200m + 192m = 8832m

Answer 2

Maximum Velocity = v = u + at

v = 0 + 0.2 x 120 = 24 ms^-1

For total distance, first do the accelerating bit-:

s = 1/2 a t^2

s = 1/2 x 0.2 x 120^2 = 1440 metres

For the constant velocity bit-:

We are going 24 ms^-1 for 300 seconds, therefore-:

24 x 300 = 7200 metres

For the retardation-:

v = u + at
0 = 24 + -1.5 x t
1.5 t = 24
t = 16 secs (to stop completely)

Now-:
s = ut + 1/2 a t^2
s = (24 x 16) + 1/2 x -1.5 x 16^2
s = 192 metres.

Total distance travelled is-:
1440 + 7200 + 192 = 8832 metres.

Answer 3

if the car was made to rest after the brakes were applied, then the answers of the 2 persons above me are correct