Question

an engine of a vehicle can produce a maximum acceleration of 4 m/s and maximum retardation of 6m/s . the vehicle is at rest initially and comes to rest after 3 km . what is the minimum time in which it can cover this distance

Answer 1

1.5 minute time distance is 3km

Answer 2

Using kinematic equation [d = V (i) t + 1 / 2 a t^2] for the first leg, we get:

D = 1/2 (5) [t (1)]^2 :(1500 - D) = [{5 t (1)} {t (2)}] + 1/2 (-10) [t (2)^2]

(1500 - D) = [{5 t (1)} {t (1)/2}] + 1/2 (-10) [{ t (1)}/2]^2

[1500 - (1/2) (5) {t (1)}^2] = [{5 t (1)} {t (1)/2}] + 1/2 (-10) [{ t (1)}/2]^2

6000 / 15 = t (1)^2 → t (1) = 20 seconds

Then t (2) = 20/2 = 10 seconds.

Total trip = 30 seconds.