1.

The acceleration a' in ms-2 of a particle is given by a = 3t^2 +2t+2, where is the time. If the particle starts with a velocity v = 2 m s-at i=0 then velocity at the end of 2 sec is:

Answer»

12`m s^(-1)`
18 `m s^(-1)`
27 `m s^(-1)`
36 `m s^(-1)`

Solution :Here `v=u+int_(0)^(t)` ADT
`=u+int_(0)^(t)(3T^(2)+2T+2)dt`
`=u+[(3t^(3))/(3)+(2t^(2))/(2)+2t]_(0)^(2)=2+8+4+4`
=18`ms^(-1)`


Discussion

No Comment Found

Related InterviewSolutions