Question

pls help to solve this question​

Answer 1

v2-u2=2as

Now, when accelerating, v = v ; u = 0; a = 5m/s2.

v2-02=2(5)s

Now when decelrating, u = v; v = 0, a = -10 m/s2

02-v2=2(-10)s

Combining both the distances should be equal to 1500m.

So, v2/20+v2/10=1500

3v2/20=1500

3v2=30000

v = 100m/s.

So, the minimum time can be found out by:

Acceleration:

v-u=at

=100-0=5t

t = 20 s

Deceleration:

v-u=at

0-100=-10t

t=10s

Total time taken is 30s.