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 it’s velocity and acceleration be perpendicular to each other?

Answer 1

Answer:

$\frac{2}{3}$

Explanation:

1)  We have,

Particle moves in the X-Y plane whose coordinates are $(t^2,t^3-2t)$

Hence, Obtaining velocity from position vector.  

$\vec{r}=(t^2,t^3-2t)\\ \\=>\frac{d\vec{r}}{dt}=(2t,3t^2-2)\\ \\=>\vec{v}=(2t,3t^2-2)$

2)Then,

differentiating velocity to get acceleration:

$\frac{d\vec{v}}{dt}=(2,6t)\\ \\ \vec{a}=(2,6t)$

3) Instant at which velocity and acceleration are perpendicular.

$\vec{a}.\vec{v}=0\\ \\=>(2,6t).(2t,3t^2-2)=0\\ \\=>4t+18t^3-12t=0\\ \\=>18t^3-8t=0\\ \\=>9t^2=4,t=0\\ \\=>t=\frac{2}{3},t=0$

Hence , at t=2/3s and t=0s , velocity and acceleration are perpendicular to each other.

Answer 2

Answer:

Given:

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

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       ………….(1)

The acceleration is,  

a= dv/dt  

 = 2i+ 6tj                …………..(2)

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

v.a= 0

Put values of eq (1) & (2)

=> (2ti+ 3t2j– 2k).(2i+ 6tj) = 0

=> 4t + 18t2= 0

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

Thus, the required time is t = 0 s.