- #1

DeanBH

- 82

- 0

height at t seconds =

y= (1/4)t^4 -26t^2 + 96t t = between 0 and 4

differentiate to find velocity = t^3 + 52t + 96

acceleration = 3t^2-52

thats part 1 of question done, easy.

Verify that y has a stationary value when t = 2 determine whether it is min/max value.

i can put this into 3t^2-52t+96 and find out that it is.

but how is it min/max

nevermind : i put it into acceleration and find out if acceleration is more or less than the value i got for velocity. If it's less i know that the thing will never accelerate again after this given time. so it would be a maximum. Where as if it was more than the value i got for velocity i know that the object would be able to accelerate again, which is minimum.. right?