Question

A body moving with uniform retardation covers 3km before its speed is reduced to half of its initial value.It comes to rest in another distance of?

Answer 1

let us suppose that its initial velocity = u

using equation of motion

v2=u2+2as

(u/2)2 = u2 =2*a*3000 (value of final velocity is haif he initial taht is u/2 and s = 3 km =3000m)

(u2/4) – u2 =2*a*3000

(-3u2/4)=2*a*3000

– u2=(2*a*3000*4)/3

u2 = – 8000*a ….............................1

by appling the same equation when it comes to rest we get

02=(u/2)2 +2*a*s

– u2/4 =2*a*s

– u2/(4*2*a) = s

– u2/8a = s

after substituting values form 1 we get

– (-8000*a)/8a = s

s = 1000m = 1km

Answer 2

S✌️OLUTION✌️✨✨✨

FIRST CASE

s=3km....v=u/2......and u=u

from equations.

..v=u-at

=>u/2=u-at

=>u=2at

now.....from equations...

s=ut -0.5×at^2

3=2at^2 -(at^2/2)

at^2=2

therefore...

ut= 2at^2=2×2=4

now second case:::;;

s'=(ut/2)-(1/2)at^2

s'=(4/2)-(1/2)2

s'=2-1=1km

so it will cover further 1 km path......