Question

A body travels a distance of 2 m in 2 s and 2.8 m in next 4 s.
What will be speed of the body at the end of 10th second
from the start.
(C.B.S.E. 2001) [Ans. 0.1 m s-1]​

Answer 1

Answer:

Explanation:

case (i) S=2m , t=2s

case(ii) S=2+2.8=4.8m  ,t=2+4=6s

let u and a be the initial velocity and uniform acceleration of the body.

we know that, S=ut+1/2 a

case(i),  2=ux2+1/2 ax or

            1=u+a     (i)

case(ii),  4.8=ux6+1/2 ax

            0.8=u+3a  (ii)

subtracting (ii) from (i), we get

           0.2=0-2a=-2a  or a=-0.2/2=-0.1m/

  put the value of 'a' in (ii) ,we get

         0.8=u+3x (-0.1)

         0.8=u-0.3

          0.8+0.3=u

         u=1.1 m/s

for the velocity of body at the end of10th second, we have

u=1.1m/s  ; a=-0.1m/s2 ,v=? ,t=10 s

as ,v=u +at

hence, v= 1.1 +(-0.1)x10=1.1-1= 0.1m/s