1.

A body of mass 2 kg is moving according to the equation for displacement at seconds as x(t) = pt^(2) + rt^(3). If p = 3 ms^(-1) , q = 4 ms^(-1) and r = 5 ms^(-1) the force acting after 2 sec is :

Answer»

`136N`
128N
68N
64N

Solution :Here `X(t)=pt+qt^(2)+rt^(2)`
`(dx)/(dt)=p+2qt+3et^(2)`
`a=(d^(2)x)/(dt^(2))=0+2q+6rt`
ACCELERATION at `t = 2s = 0 + 2 xx 4 + 6 xx 5 xx 2`
`= 68 ms^(-2)`
`"Force" = ma = (2 xx 68) N`
`= 136 N`
HENCE .a. is the right choice


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