Question

a particle moving with a constant acceleration describes in the last second of its motion 36% of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it moves if it describes 6cm in the first sec.?

Answer 1

Assume the particle's acceleration is a cm/sec^2 and that it moves a distance of S(t)cms in total time of t secs. Also it covers S(t-1) cms. in (t-1) secs. 

Since it covers 6cm in the first second => 6 = 0.1 + (1/2).a.1^2 => a=12cm/sec^2. 

Now, S(t)-S(t-1)= (1/2).a.[t^2-(t-1)^2]......(1) 

Also, S(t)-S(t-1)= (9/25).S(t)= (9/25).(1/2).a.t^2 .....(2) 
From the above two equations, you will get the following eqn. 
9t^2-50t+25 = 0 
Solve it to get two values for t i.e. t=5,10/18 (ignore the second as t is not less than 1sec) 
Thus t=5sec and hence S(t)=150cm 
Therefore,option (1) is correct.