Question

A particle along a straight path ABC with a uniform acceleration of 0.5 m/s² while it crosses A, its velocity is 5 m/s. It reaches with velocity of 40 m/s, 30 s after it has crossed B in its path. Find the distance AB.​

Answer 1

Answer:

The distance AB= 600m

Explanation:

Given Data:

Initial velocity, u= 5m/s

Final velocity, v= 40m/s

Acceleration, a= 0.5m/s^2

let the time being taken be t.

v= u+at

40= 5+(0.5)t

t= 70 sec

Time taken to cross B point= t-30= 40 sec

Now,

AB= s= ut+1/2at^2= (5×40)+1/2×(0.5)×(40)^2= 600m

Therefore, SOLVED!

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