A scooter can produce a maximum acceletation of 5m/s^2.it's brakes can produce a maximum retardation for 10m/s^2.find the minimum time in which it starts from rest covers a distance of 1.5km and stops again?

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Solution

v²-u²=2as
Now, when accelerating, v = v ; u = 0; a = 5m/s².
v²-0²=2(5)s
s=v^2/10
Now when decelrating, u = v; v = 0, a = -10 m/s²
0²-v²=2(-10)s
s=v^2/20
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.

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