Question

A particle starts from origin at `t=0` with a constant velocity `5hatj m//s` and moves in `x-y` plane under action of a force which produce a constant acceleration of `(3hati+2hatj)m//s^(2)` the `y`-coordinate of the particle at the instant its `x` co-ordinate is `24 m` in `m` is
A. 12 m
B. 24 m
C. 36 m
D. 48 m

Answer 1

Correct Answer - C
The position of the particle is given by
`vecr = vecr_(0) = vecv_(0)t + (1)/(2) vecat^(2)`1
where, `vecr_(0)` is the position vector at t = 0 and
`vecv_(0)` is the velocity at t = 0
Here, `vecr_(0) = 0, vecv_(0) = 5hati ms^(-1 ), veca = (3hati + 2hatj) ms^(-2)`
`therefore vecr = 5thati +(1)/(2)(3hati +2hatj) t^(2) = (5t + 1.5t^(2))hati + 1t^(2)hatj " "..(i)`
Compare it with `vecr = x hati + y hatj,` we get
`x = 5t + 1.5t^(2), y = 1t^(2)`
`because x = 84 m`
`therefore 84 = 5t + 1.5t^(2)`
On solving,we get t = 6 s
At t = 6 s, y = (1) `(6)^(2) = 36` m