Question

The maximum acceleration attained on the interval [0,3] by the particle whose velocity is given by v(t)=t^3-3t^2+12t+4

Answer 1

when I took the derivative of v'(t) and tried to solve v'(t)=0 my solutions were complex so I am only testing 0 and 3 into v(t) correct?

Answer 2

I know the answer

Answer 3

@sourwing

Answer 4

you're trying to maximize a(t). So you need to set a'(t) = 0

but a(t) = v'(t), and so a'(t) = v''(t).

Answer 5

so what you should have done is v''(t) = 0. Not v'(t) = 0

Answer 6

solve and t = 1.
determine the largest value among a(0) , a(1) and a(3)

Answer 7

but 12 is not the solution

Answer 8

a(3)=12 is not the solution

Answer 9

21 is my solution and that would be v(3)=21

Answer 10

I thought this was an absolute max problem

Answer 11

a(3) = 21. check your math

Answer 12

now I am confused so let me write out my work

Answer 13

v(t) is the velocity
a(t) is the acceleration and the first derivative of v(t)
correct

Answer 14

yes

Answer 15

v(t)=t^3-3t^2+12t+4
a(t)=v'(t)=3t^2-6t+12

Answer 16

yes, and you're maximizing a(t)

Answer 17

all know about particle motion is PVA
position, velocity and acceleration

Answer 18

no one ever taught me to take the derivative of A

Answer 19

but you are saying to maximize "something", you need to take the derivative of "something"

Answer 20

yes, you're trying to maximize *acceleration*, so you need to set a'(t) = 0

Answer 21

ok thanks.

Answer 22

so maximum value of acceleration is 21 which occurs at t = 3

Answer 23

yes, thanks once again.