Question

A particle moving with constant acceleration describes in the last second of its motion 9/25th of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it move if it describes 6 cm in first second.

Answer 1

Answer :-

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

s = ut+{1/2} a{t}^{2}

where

t=1sec

s=6cm

u=0

s = ut +{1/2} a{t}^{2}

6 = (1/2)a1

6*2=a

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)at^2 .....(2)

form 1and 2 we get

9{t}^{2} - 50t+25 = 0

9t^2- 50t+25 = 0

9t^2- 50t+25 = 0

t=5

t =5/9 (not possible)

v= u+at

where ,u=0 ,t=5 ,a=12

v=12t

v=12tv=12*5

v=12tv=12*5

v=60

{v}^{2} = {u}^{2} + 2as

{60}^{2} = 0 + 2*12 s

3600 = 2*12 s

3600/24= s

24= ss = 150

distance =150cm

distance =150cmtime=5

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