Question

A body moving with uniform retardation covers 3km before its speed is reduced to half of its initial value.

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) \therefore (u2/4) – u2 =2*a*3000 \therefore (-3u2/4)=2*a*3000 \therefore – u2=(2*a*3000*4)/3 \therefore u2 = – 8000*a                      ….............................1 by appling the same equation when it comes to rest we get      02=(u/2)2 +2*a*s \therefore – u2/4 =2*a*s \therefore – u2/(4*2*a) = s \therefore – u2/8a = s after substituting values form 1 we get \therefore – (-8000*a)/8a = s \therefore s = 1000m = 1km i think this might help