Question

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.

Answer 1

Answer:

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

$v(t) = s'(t) = 3t^2 - 8$

The acceleration function si the derivative of velocity function

$a(t) = v'(t) = 6t$

a. When s(t) = 0 then

$t^3 - 8t = 0$

$t(t^2 - 8) = 0$

t = 0 or

$t^2 - 8 = 0$

$t = \sqrt{8} = 2.83 s$

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

$v(0) = -8$ so negative direction

$v(\sqrt{8}) = 3*8 - 8 = 16$ 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

$v(t) > 0$

$3t^2 - 8 > 0$

$t^2 > 8/3$

$t > \sqrt{8/3} = 1.633 s$ or $t < -\sqrt{8/3} = -1.633$ (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