1.

A body travels a distance of `2 m` in `2` seconds and `2.2m` next `4 seconds`. What will be the velocity of the body at the end of `7` the second from the start?

Answer» Here, case 9i) `S=2 m, `t=2 s`
Case (ii) `S=2 +22.2 =42 , t=2=6 s`
Let (u) and (a) be the initial velocity and unitorm acceleration of the body.
we know that, `S=ut + 1/2 at^(2)`
Case 9i), `2=u xx 2 + 1/2 a xx 2^(2)`
or `1=u+a` .....(i)
Case (ii) , `4.2 =u xx 6 +1/2 axx 6^(2)`
or `0.7 =u +3a` ......(ii)
Subtracting (ii) form(i), we ger
`0.2 =0-2a =- 2a`
or `a =- 0- 0.3//2 =- 0.15 ms^(-20`
From (i0, `u=1 -a =1 +0.15 = 1.15 ms^(-1)`
For the velcity of body at the end of `7 thsecond, we have `u=1.15 ms^(-1)` ,
`a=- 0. 15 ms^(-2) , v-? t=7 s`
As, v=u +at `
:. `v=1.15 +(-0.15) xx 7 =0.1 ms^(-1)` .


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