Question

Given an acceleration vector, initial velocity u0,v0,w0 , and initial position x0,y0,z0 , find the velocity and position vectors for t ge 0. a(t) = 5,5t,8t , u0,v0,w0 = 15,0,0 , x0,y0,z0 = 0,0,0 What is the velocity vector?

Answer 1

Answer:

Thus we find that velocity vector at time t is

(5t+15, 5t^2/2, 4t^2)

Step-by-step explanation:

given that acceleration vector is a funciton of time and at time t

$a(t) = (5,5t, 8t)$

v(t) can be obtained by integrating a(t)

v(t) = $(5t, 5t^2/2, 4t^2)+(u_0,v_0,w_0)\\=(5t+15, 5t^2/2, 4t^2)$

Thus we use the fact that acceleration is derivative of velocity and velocity is antiderivative of acceleration.

The arbitary constant normally used for integration C is here C vector = initial velocity (u0,v0,w0)

Position vector can be obtained by integrating v(t)

Thus we find that velocity vector at time t is

(5t+15, 5t^2/2, 4t^2)