Question

If an object moves in a straight line in such a way that its distance (in meters) from a given point at time t (in seconds) is given by s(t)=2t^3-9t^2+24.

a) Find the velocity and acceleration at time t = 2 seconds.

b) Find the velocity at the time(s) when the acceleration is zero.

c) Find the acceleration at the time(s) when the velocity is zero.

Answer 1

for (a) i found the derivative and placed 2 for t, but i ended up with a -12, which doesn't make sense to me.
and that's only the velocity.. not sure how to find the acceleration.

Answer 2

A negative velocity means it is moving in the negative direction (whatever is opposite to what you are defining as the positive direction).

Answer 3

Velocity is a vector, so it includes direction as well as magnitude.

Answer 4

Acceleration is the derivative of velocity.

Answer 5

so to do my velocity, i get the derivative of the function and sub in 2.
to get the acceleration, i just get the derivative of the velocitiy, which is the derivative of the function?

Answer 6

Yes. Velocity is the first derivative of position, and acceleration is the first derivative of velocity; thus, acceleration is the second derivative of position.

Answer 7

so for part b, do i put the first derivative or the second derivative equal to zero?

Answer 8

Second. Accel.=0 -> s''=0.

Answer 9

so i set it equal to zero, i dont sub in zero?

Answer 10

No, you are solving for t when s''(t)=0.

Answer 11

im doing part b now.

Answer 12

Ok, let me know what you get.

Answer 13

so i do 12t-18=0?

Answer 14

t=3/2

Answer 15

Correct.

Answer 16

okay so for part c, do i sub in zero into my second derivative?

Answer 17

No, set s'(t)=0 and solve for t.

Answer 18

*remember, when you are solving for t, you are getting the time. The questions are asking for the velocity (b) and the acceleration (c) at those times.

Answer 19

=0

Answer 20

?

Answer 21

for c, should my answer be 0?

Answer 22

No, and for that matter, what was your final answer for (b)?