The position function s(t)=t3−8t gives the position in miles of a freight train where east is the positive direction and t is measured in hours.

a. Determine the direction the train is traveling when s(t)=0 .
b. Determine the direction the train is traveling when a(t)=0 .
c. Determine the time intervals when the train is slowing down or speed up.

a. Negative direction when t = 0s and positive direction when t = 2.83s

b. Negative direction

c. The train would be slowing down when t is between 0 and 1.633 s then speeding up when t is > 1.633 s

Step-by-step explanation:

The velocity function is the derivative of position function

The acceleration function si the derivative of velocity function

a. When s(t) = 0 then

t = 0 or

Plug both of the ts into the velocity function and we have

so negative direction

so positive direction

b. When a(t) = 0 then 6t = 0 so t = 0. v(0) = -8 so negative direction

c. As a = 6t and t is larger or equal to 0. Then a is also larger or equal to 0. The trains is speeding up if v is positive and slowing down when v is negative

When v is positive

or (not possible)

Similarly, v is negative when t < 1.633 s

So the train would be slowing down when t is between 0 and 1.633 s then speeding up when t is > 1.633 s

5050 i think

step-by-step explanation:

thats not a system of equations, that is a solution