The position of an object moving horizontally after t seconds is given by the function s equals 3 t minus t cubed , for t greater than 0 , where s is measured in feet, with s greater than 0 corresponding to positions right of the origin.

a. When is the object stationary, moving to the right, and moving to the left?

b. Determine the velocity and acceleration of the object at tequals2.

c. Determine the acceleration of the object when its velocity is zero.

d. On what intervals is the speed decreasing?

Step-by-step explanation:

The position of an object moving horizontally after t seconds is given by the function

s = 3t - t³

a) The object is stationary when there is no external force acting on the body. When the body is at rest, the body remains in a position and here is no distance covered by the object i. e s = 0

b) velocity is the change in displacement of a body with respect to time.

v = ds/dt

S = 3t - t³

V = ds/dt = 3-3t²

at t = 2

Velocity = 3-3 (2) ²

Velocity = 3-12

Velocity = - 9m/s

c) acceleration is the change in velocity of a body with respect to time.

acceleration = dv/dt

If v = 3-3t²

a = dv/dt = - 6t

When v = 0

0 = 3-3t²

-3 = - 3t²

t² = 1

t = ±√1

t = 1sec

The acceleration of the object at v = 0 occurs at t = 1sec and - 1sec

a = - 6 (1)

a = - 6m/s²

d) Given the speed of the body v modelled by the function

v = 3-3t²

The speed is decreasing when it is less than zero as shown:

3-3t²< 0

3<3t²

1
±1
1
t>±1

t >1 and t>-1

The speed is decreasing when

3-3t²1 or t>-1