Consider a toy car that moves back and forth on a long, straight track for 4 minutes in such a way that the function s(t)=96t-84t^2+28t^3-3t^4 describes how far the car is to the right of the starting point, in centimeters, after t minutes.

When is the toy car accelerating the fastest to the right? When is the toy car accelerating fastest to the left?

Question

Consider a toy car that moves back and forth on a long, straight track for 4 minutes in such a way that the function s(t)=96t-84t^2+28t^3-3t^4 describes how far the car is to the right of the starting point, in centimeters, after t minutes.

When is the toy car accelerating the fastest to the right? When is the toy car accelerating fastest to the left?

anonymous · Answer

I am having a hard time with trying to optimize this problem :(

anonymous · Answer

translate this as 
"find the maximum value of the second derivative of $[0,4]$

anonymous · Answer

*of
and i guess also the minimum of the second derivative

anonymous · Answer

i get $$-12 (3 t^2-14 t+14)$$

anonymous · Answer

Ok so what I have so far is the first, second, and third derivative of the equation and I calculated t=2.33 for the third derivative. I then put that number into the second derivative as well as 0 and 4 and I had 2.33 being the most positive and 0 being the most negative otu of the three values.

anonymous · Answer

max is at the vertex 
min is at 0 i think

anonymous · Answer

so do i need the third derivative at all?

anonymous · Answer

yeah you got 0 as the most negative, so that looks good
the vertex is $\frac{14}{6}=\frac{7}{3}$
you can use it if you like, but you don't need calculus to find the max and min or a quadratic
check the endpoints and the vertex

anonymous · Answer

of course you will get the vertex by taking the derivative, setting it equal to zero and solve 
$$f(t)=at^2+bt+c$$
$$f'(t)=2at+b$$
$$2at+b=0\iff t=-\frac{b}{2a}$$ same thing

anonymous · Answer

ok let me just change it quickly...

anonymous · Answer

no need, i think you got the same answer, just used a decimal instead of my fraction

anonymous · Answer

oh i erased my work already :( this is a little easier for me though

anonymous · Answer

since $\frac{7}{3}=2\frac{1}{3}=2.333...$

anonymous · Answer

the first derivative is so long :( i have so far...

t=8+7t^2-t^3/4

anonymous · Answer

Ok I got it back so I have t=2.33 having the most positive s''(t) value and t=4 having the most negative value

anonymous · Answer

sorry i mean t=0 as most negative

anonymous · Answer

so that would mean...the car accelerates the fastest to the right at t=2.33 and the car accelerates fastest to the left at t=0?

anonymous · Answer

or at least i think it is right lol