Question

the position of a particle moving along a horizontal line is given by x(t)=4 (t-4)^3. What is the maximum speed of the particle for 0 14 years ago

Answer 1

find the max of the derivative

Answer 2

you got this?

Answer 3

would the derivative be x'(t)=12 (t-4)^2

Answer 4

that's right..

Answer 5

v(t)=x'(t)=12 (t-4)^2

Answer 6

this is a parabola that faces up so it doesn not have a maximum
however, your interval is \(0\leq t\leq 10\) so you only have to check at the endpoints for a max

Answer 7

t=10
v=432 m/s