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.
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
A train starts from rest at a station and accelerates at a rate of 1.5 m/sec2 for 16 seconds
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