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.
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A particle moving with a constant acceleration describes in the last second of it's motion 9/25th of the whole distance.of it starts from rest ,how long is the particle is in motion and through what distance does it move if it describes 6cm in first second?
options
1. 5s,150 cm
2. 10s,150 cm
1 answer
1 answer