Question

A car accelerates from rest at a constant rate 'alpha' for some time after which it decelerates at a constant rate 'beta' and comes to rest. if total time elapsed is 't', then max velocity acquired by the car is

Answer 1

Answer:

Explanation:

For the first leg of the journey,

V = 0 + α  t1

So  V = α  t1

for the second leg of journey of duration  t - t1 :

0 = V - β (t - t1)

0 = α t1 - β t + β t1

t1 = β t / (α + β)

V =  αt1 = α β t / (α + β )

Answer 2

Answer:

Explanation:

For the first leg of the journey,

V = 0 + α  t1

So  V = α  t1

for the second leg of journey of duration  t - t1 :

0 = V - β (t - t1)

0 = α t1 - β t + β t1

t1 = β t / (α + β)

V =  αt1 = α β t / (α + β )