Question

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?

Answer 1

Answer:

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<t²

±1<t

1<t and -1<t

t>±1

t >1 and t>-1

The speed is decreasing when

3-3t²<0 and t>1 or t>-1