A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec ?

Options

A. At `t=2//3 s`, directions of velocity and acceleration are perpendicular
B. At `t=0`, directions of velocity and acceleration are perpendicular
C. At `t=sqrt(2/3) s`, particle is moving parallel to x-axis
D. Acceleration of the particle when it is at point `(4, 4)` is `2hat(i)+24 hat(j)`

Answer

A, B, C, D

Explanation

`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)`