The maximum acceleration attained on the interval zero which is less than or equal to t which is less than or equal to 3 by the particle whose velocity is given by v(t)=t^3-3t^2+12t+4 is 21. But I don't understand how to solve it. Someone please explain this to me.

Question

ddcamp · Answer

Is this a calculus question?
If so, we know that acceleration is the derivative of velocity. So,
a = 3t²-6t+12
Given 0≤t≤3, what is the maximum value of a?

anonymous · Answer

Yeah, it is a calculus problem XD But I don't understand how to find the maximum value of a. I tried to solve it like I would to find relative maximums by setting 3t^2-6t+12=0, but I keep getting imaginary answers. What should I do?

ranga · Answer

No to find the maximum/minimum acceleration you need to find its derivative and set it to 0 and solve for t.

ddcamp · Answer

Did you learn in pre-calc that, with a quadratic functions, a maximum or minimum of the function
y = ax²+ bx + c
is always located at x = -b/2a?
If not, that will be useful to remember.

anonymous · Answer

OHMYGOD I JUST GOT IT. Wow. I feel really stupid now... Haha thanks!

ranga · Answer

you are welcome.