Question

A partical moves with uniform acceleration travels 24 m and 64m in first two consecutive intervals of 4sec .What is the initial vilocity

Answer 1

24=u*4+a*4*4/2

4u+8a=24

u+2a=6. (i)

64=u*8+a*8*8/2

8u+32a=64

u+4a=8. (ii)

Solve it simultaneously for your own practice

Answer 2

let initial velocity u = m/s

let acceleration a=m/s^2

• let distance travelled in first first 4s interval is S1,

it is given by,

S1=u×t+1/2×a×t^2..........(1)

• from 1 we get,u+2×a=6.......(2)

after 4s , velocity =a+u×t =u +4a

distance S2 travelled by 4s interval is given by,

S2=(u+4×a)×t+1/2×a×t^2.......(3)

from 3 we get u+6a=16........(4)

by solving (2) and (4) we get

u=1m/s

a=2.5m/s^2