Question

A train starts from rest at a station and accelerates at a rate of 1.5 m/sec² for 16 seconds. It runs at a constant speed for 80 seconds and slow down at a rate of 2.5 m/sec² until it stops at next station. Find the total distance covered by train.

Answer 1

Divide the problem in 3 parts, First accleration from rest, second constant speed , third deceleration to rest .

Ist part,

Let , initial speed = u = 0m/s (since train starts from rest)

acceleration = a = 1.5 m/s^2

time taken = t = 16 s

Therefore, final velocity v at the end of first part of motion ,v = u + at

v=u+at

=0 + 1.5 * 16 = 24 m/s

Let s1 be distance travelled in first part of motion,

Another equation, v^2 = u^2 + 2 *a *s1

24^2 = 0 + 2 * 1.5 * s1

576 = 3 * s1

s1= 576/3 m ....... [ 1st ]

2nd part

The train travels with constant speed 24m/s [ final speed of first part ] for 80 seconds,

Let distance travelled for 2nd part be s2

s2 = 24 * 80 = 1920m ... [ 2nd ]

3rd part

Initial velocity of 3rd part is Final velocity of 2nd part ( which is constant = 24m/s )

Initial velocity = u = 24 m/s

Final velocity of 3rd part = v = 0 ( since train comes to rest )

accleration = a = - 2.5 m/s^2

Let distance travelled in 3rd part be s3

By using equation, v^2 = u^2 + 2(-a) s3

0 = 24^2 - 2 * 2.5 * s3

5 * s3 = 576

s3 = 576/5 .... [ 3rd ]

Total distance travelled = s1+s2+s3 = 576/3 + 1920 + 576/5 = 2227.2 m