a train starts from rest and acquires a speed v with uniform accelerationalpha . then it comes to a stop with uniform retardation beta what will be the average velocity of the train of answer is v/2

Hello!

Let us consider the first statement. The train starts from rest and acquires a speed v. So, initial velocity is 0 and final velocity is v.

We know that for uniform motion,

v = u + at. Here a = α .

Therefore,

v = u + α t

Since u = 0,

v = α t

α = v/t

Let t = t1

Thus

α  = v/t1 ..........(1)

Also, v^2 - u^2 = 2as

Since a = α, u = 0 and s = s1 is the distance travelled

v^2 = 2 α s1 ...........(2)

From (1) and (2), equating the values of  α, we get

s1 = v t1/2 ............(I)

Before it comes to stop,

v = u + at

Since it comes to stop, v =0; also here, a = β

Thus

0 = u + βt

Putting t = t2

0 = u + β t2

Now, the final velocity when the train starts from rest and the initial velocity before the train comes to rest are the same.

That is, u = v

Therefore,

0 = v + β t2

β = -v/t2 .............(3)

Also, v^2 - u^2 = 2as

Since a = β, final velocity v = 0 (as the train comes to a stop) and s = s2 is the distance travelled,

0 - u^2 = 2 β s2

We know that u = v. Thus,

-v^2 = 2 β s2 ........(4)

From (3) and (4), equating the values of β , we get

s2 = v t2/2 ............(II)

The average velocity can be calculated by

Avg. Velo. =  Total Distance Travelled/Total Time Taken

=  s1+s2/t1+t2

Substituting the values of s1 and s2 from (I) and (II),

Avg. Velo. =  (vt1/2 + vt2/2) / t1+t2

=   v/2 (t1+t2) /(t1+t2)

= v/2

Thus, the average velocity is v/2 .