Question

The equation r(t)=(8 sin t) i+(8 cos t) j+(8t) k is the position of a particle in space at time t. Find the​ particle's velocity and acceleration vectors. Then write the​ particle's velocity at t= 3π 2 as a product of its speed and direction.

Answer 1

Answer:

Velocity is $8\sqrt{2}$ and Acceleration is 8

Step-by-step explanation:

Equation of position of the particle

r(t) = (8sint)i + (8cost)j +  (8t)k

1) Velocity of the particle is $\frac{dr}{dt}$

 V(t) =  dr/dt = (8cost)i + (-8sint)j + 8k

2) Acceleration of the particle will be $\frac{dv}{dt}$

    A(t) = dv/dt = (-8sint)i + (-8cost)j + 0k

velocity of the particle at t= $\frac{3pi}{2}$

 V($\frac{3pi}{2}$) = (0)i + (8)j + 8k = $\sqrt{8^2+8^2}$ = $8\sqrt{2}$

acceleration at t= $\frac{3pi}{2}$

A($\frac{3pi}{2}$) = (8)i + (0)j +0k = $\sqrt{8^2}$ = 8