Problem: Find the average speed of a train that takes t sec to perform a journey. It travels for t/n sec with uniform acceleration, then for (n-3)t/n sec with uniform speed v and finally comes to rest with uniform retardation.

Solution:

To find the average speed of the train, we need to find the total distance covered by the train and divide it by the total time taken.

Let's assume that the initial velocity of the train is u and the final velocity is zero.

Step 1: Finding the distance covered during the period of acceleration

Using the equation of motion, we can find the distance covered during the period of acceleration.

v = u + at

where,
v = final velocity (which is constant during uniform speed)
u = initial velocity (which is zero in this case)
a = acceleration (which is constant during the period of acceleration)
t = time taken for acceleration (which is t/n)

So, the final velocity after the period of acceleration is:

v1 = u + at
v1 = 0 + a(t/n)
v1 = at/n

The average velocity during the period of acceleration is:

(av)1 = (u+v1)/2
(av)1 = (0+at/n)/2
(av)1 = at/2n

The distance covered during the period of acceleration is:

s1 = ut + 1/2at^2
s1 = 0(t/n) + 1/2a(t/n)^2
s1 = 1/2at^2/n^2

Step 2: Finding the distance covered during the period of uniform speed

During the period of uniform speed, the train travels at a constant speed v for (n-3)t/n sec. So, the distance covered during this period is:

s2 = v(n-3)t/n

Step 3: Finding the distance covered during the period of retardation

During the period of retardation, the train starts with a constant speed v and comes to rest at the end of the journey. So, the initial velocity is v and the final velocity is zero. Using the equation of motion, we can find the distance covered during the period of retardation.

v^2 = u^2 + 2as

where,
v = final velocity (which is zero)
u = initial velocity (which is v)
a = retardation (which is constant during the period of retardation)
s = distance covered during the period of retardation

So, the distance covered during the period of retardation is:

s3 = v^2/2a
s3 = v^2/2(-a)
s3 = v^2/(-2a)

Step 4: Finding the total distance covered

The total distance covered by the train is the sum of the distances covered during the three periods.

s = s1 + s2 + s3
s = 1/2at^2/n^2 + v(n-3)t/n + v^2/(-2a)

Step 5: Finding the average speed

The total time taken by the train is t. So, the average speed is:

average speed = total distance / total time
average speed = s/t
average speed = [

Ans:- (2n-3) (v) /(2n )