1.

A body travels 200 cm in the first two seconds and 220 cm in the next 4 seconds with deceleration. the velocity of the body at the end of the `7^(th)` second is

Answer» Correct Answer - `10cms^(-1)`
Let `u` be initial velocity `&a ` be its acceleration
Distance in first `2 sec =S_(1)=200cm`
`rArr u(2)+(1)/(2)a(2)^(2)=200cm`
`rArr u+a=100` ...`(i)`
Distance in next `4 sec. =S_(2)=220cm`
Distance in first `6 sec. =S_(1)+S_(2)=200+220cm`
`rArru(6)+(1)/(2)a(6)^(2)=420`
`rArr u+3a=70` ....`(ii)`
From equations `(i) & (ii), ` we get
`a=-15cm//s^(2),u=115 cm//s`
Hence, velocity at the end of `7 sec.` from start
`=u+7a`
`=115+7(-15)=10cm//s`.


Discussion

No Comment Found

Related InterviewSolutions