InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A particle moves in the xy plane and at time t is at the point `(t^(2), t^(3)-2t)`. Then :-A. At `t=2//3 s`, directions of velocity and acceleration are perpendicularB. At `t=0`, directions of velocity and acceleration are perpendicularC. At `t=sqrt(2/3) s`, particle is moving parallel to x-axisD. Acceleration of the particle when it is at point `(4, 4)` is `2hat(i)+24 hat(j)` |
|
Answer» Correct Answer - A::B::C::D `vec(r)=t^(2)hat(i)+(t^(3)-2t)hat(j)`, `vec(v)=(dvec(r))/(dt)=2that(i)+(3t^(2)-2)hat(j)` `vec(a)=(d^(2)vec(r))/(dt^(2))=2hat(i)+6that(j)` `vec(a).vec(v)=4t+18t^(3)-12t=0 ("For " bot)` `:. T=+- 2//3, 0`. For parallel to x-axis `rArr (dy)/(dx)=0 rArr (dy)/(dx)=(3t^(2)-2)/(2)` `:. at t=sqrt(2/3)` sec it becomes zero so (c) `vec(a)(4, 4)=2hat(i)+6xx2hat(j)=2hat(i)+12 hat(j)` |
|