Is Uniform Velocity that easy ?
A body starts its motion from rest and accelerates with a = 5m/s^2 then moves with uniform velocity and finally decelerates with a = 5m/s^2.The total time for the journey is 25 s and the average velocity is 72 km/hr. Find the time the body travelled in uniform motion.
The answer is 15.
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2 solutions
nice solution...
Let the time for acc. and dec. each be t. So the constant speed reached is 5t. 72mil/hr=20m/s. Total distance travelled is 20m/s * 25s=500.
25-2t is the distance travelled with constant speed of 5t.
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Just a different approach.
When Accelerating..
s = 1/2 x 5 x t^2
When Decelerating..
s = 5 x t^2 - 1/2 x 5 x t^2 = 1/2 x 5 x t^2 (Since from v = u + at, v = 5t as u = 0)
Total Distance travelled when the body was in non-uniform motion = 5t^2
While travelling in uniform motion..
s = (25 - 2t) x 5t
= 125t - 10t^2
Total Distance Travelled = Time x Average Velocity (72 km/h = 20m/sec)
125 - 10t^2 + 5t^2 = 25 x 20
t^2 - 25t + 100 = 0
t = 5,20
Since t = 20 is impractical, value of t = 5
Hence the time the body travelled in uniform motion = 25 - 2t = 15
Hey Hrishik Mukherjee your solution is nice . BTW ya' should learn LATEX :)
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I will learn it later bro..
This is velocity time graph of the situation.
From the graph. Total distance traveled till 25 seconds is are of the trapezium
So total distance
= 2 1 ( 5 t 0 ) ( 2 5 + 2 5 − 2 t 0 )
As A v . S p e e d = T o t a l _ t i m e T a t a l _ D i s t a n c e
Using this I got total distance traveled=500m
So
5 0 0 = 2 1 ( 5 t 0 ) ( 2 5 + 2 5 − 2 t 0 )
on solving we get t 0 = 5
So the time the body travelled in uniform motion is 1 5 s e c