Question

Derivatives as a rate of change:
s=(t^4/4)-t^3+t^2, (interval greater than or equal to 0 and less than or equal to 3.
a.) find displacement and average velocity for the given time interval.
I said: f(3)-f(0)=2.25 and avg velocity= -3/4...is that right?
b.) find the speed and acceleration at the endpoints of the interval. I said: s'=t^3-3t^2+2t. s''=3t^2-6t+2. |v(0)|=0; |v(3)|= 6.
c.) when, if ever, during the interval does it change direction?
???? I don't know what this means...

Answer 1

when does the acceleration equal 0? s = distance, ds/dt = velocity, d^2s/dt^2=acceleration. When the acceleration is 0, the velocity will change signs from positive to negative, or vice versa.

Remember in calc 1, you took the 2nd derivative of a parabola and set it equal to zero to find when the curve's vertex, where it "turned"

Answer 2

hmm...I'm in calc 1 now and I honestly don't remember doing that.

Answer 3

Your s'(t) is incorrect. s'(t)=4t^3-3t^2+2t, you're missing the 4 coefficient in front of the t^3.

Perhaps you haven't learned that yet in calc 1, or you're doing what I did when I took calculus and slept through all the lessons.

What you want to do is set s''(t) = 0, solve for the roots of that equation. If that fits in your domain (0 to 3 sec) then the object changed direction during that time.

Answer 4

My professor said something about the vertex last week, but told us not to worry about that.

Answer 5

So the first derivative is s'=4t^3-3t^2+2t?

Answer 6

yes

Answer 7

so s"=12t^2-6t+2?

Answer 8

and then I set that to 0 and check if it's in my domain?

Answer 9

Thank you so much btw!