Question

A body moving with uniform retardation covers 3 km 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

Answer:

The body comes to rest in the distance of one kilometer after its speed reduced to half of initial value.

Explanation:

Given that the distance covered by body, S $=3Km$

Consider that the initial velocity $=v$

Final velocity of the body $=\frac{v}{2}$

 ∴         $v^{2}-u^{2}=2aS\\( \frac{v}{2})^{2} -v^{2}=2a(3)\\ \frac{-3v^{2}}{4}=6a\\ a=\frac{-v^{2}}{8}$

Now if initial velocity of the body $=\frac{v}{2}$

Given  the final velocity, v=0

Then      $(0)^{2}-(\frac{v}{2})^{2} =2*(\frac{-v^{2}}{8})*x$

                 $\frac{-v^{2}}{4} =\frac{-v^{2}}{4} *x\\ x=1Km$

Therefore the distance is covered is one kilometer before body comes to rest.