Consider the following position function. Find (a) the velocity and the speed of the object and (b) the acceleration of the object.

r (t) = (9, t^3, e^-t), for t greater than or equal to 0

Answer: v = 3t^2, - e^-t;

Speed = √ 9t^4 + (-e^-t) ^2;

a (t) = 6t, e^-t

Step-by-step explanation:

Given the position function at time 't' for t greater than or equal to 0:

r (t) = (9, t^3, e^-t)

The Velocity of 'v' is the first derivative of r with respect to 't':

d/dt (9) = 0

d/dt (t^3) = 3t^3-1 = 3t^2

y = (e^-t) =

Using chain rule:

Let u = - t

y = e^u; dy/du = e^u

du/dt = - 1;

Therefore dy/dt = dy/du * du/dt

dy/dt = - 1 * e^u; - e^u; u = - t

dy/dt = - e^-t

Therefore,

v = 3t^2, - e^-t

The magnitude of velocity 'v ' = speed

Speed = √v1^2 + v2^2

Speed = √ (3t^2) ^2 + (-e^-t) ^2

Speed = √ 9t^4 + (-e^-t) ^2

Acceleration (a) = derivative of velocity

d/dt (3t^2) = 6t

d/dt (-e^-t):

Using chain rule:

dy/dt = - e^-t

Let u = - t; du/dt = - 1

-e^u; dy/du = - e^u

Therefore, dy/dt = - 1 * - e^u

u = - t

dy/dt = e^-t

Therefore, acceleration:

a (t) = 6t, e^-t