Question

A particle moves in the xy plane and at time t is at the point (t^2, t^3 - 2t). Th
(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 = v
s, particle is moving parallel to x-axis
(D) Acceleration of the particle when it is at point (4, 4) is 2î + 12
An object may have :
(A) Varying speed without having varying acceleration
B Varying velocity without having varying speed
C) Non-zero acceleration without having varying velocity
D Non-zero acceleration without having varying speed,​

Answer 1

Answer:

The coordinate of the particle is = (t2, t3, -2t)

the required time is t = 0 s.

Explanation:

The position vector of the particle is = t2i+ t3j– 2tk= r(say)

The velocity of the particle is, v= dr/dt = 2ti+ 3t2j– 2k

The acceleration is, a= dv/dt = 2i+ 6tj

At time 't' the v and a are perpendicular to each other.

Therefore,

v.a= 0  

=> (2ti+ 3t2j– 2k).(2i+ 6tj) = 0

=> 4t + 18t2= 0

=> t = 0, -2/9 [t = -2/9 is absurd]

Thus, required time is t = 0 s.