A particle moves in the x-y plane and at time t is at the point whose coordinates are (t2,t3,-2t) - Physics - Motion In A Plane

Question

A particle moves in the x-y plane and at time t is at the point
whose coordinates are (t2,t3,-2t) Then at what instant of time will
its velocity and acceleration vector sbe perpendicular to each
other?

Somnath physics · Accepted Answer

The coordinate of the particle is =
(t2,
t3,
-2t)
The
position vector of the particle is =
t2i+
t3jâ€“
2tk=
r(say)
The
velocity of the particle is,
v=
dr/dt =
2ti+
3t2jâ€“
2k
The
acceleration is, a=
dv/dt =
2i+
6tj
At time
't' the vand
aare
perpendicular to each other
Therefore,

v
a=
0
=>
(2ti+
3t2jâ€“
2k)
(2i+
6tj)
= 0
=> 4t
+ 18t2=
0
=> t =
0, -2/9 [t = -2/9 is absurd]
Thus, the
required time is t = 0 s

A particle moves in the x-y plane and at time t is at the point whose coordinates are (t2,t3,-2t). Then at what instant of time will its velocity and acceleration vector sbe perpendicular to each other?