InterviewSolution
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InterviewSolution
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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` isA. 12 mB. 24 mC. 36 mD. 48 m |
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Answer» 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 |
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