Question

a particle moving with a constant acceleration describes in the last second of its motion 9/25 th 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 6cm in first sec​

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

$s = ut + \frac{1}{2} a{t}^{2}$

where

t=1sec

s=6cm

u=0

$s = ut + \frac{1}{2} a{t}^{2}$

6 = (1/2)a1

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)

From 1and 2 we get

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

t=5

$t = \frac{5}{9}$

t =5/9 (not possible)

v= u+at

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

v=12t

v=12*5

v=60

${v}^{2} = {u}^{2} + 2a {s}$ ${60}^{2} = 0 + 2 \times 12 s$

$3600 = 2 \times 12 s \\ \frac{3600}{24} = s$

${s} = 150$

distance =150cm

time=5