Question

The position (in feet) of an object is given by s(t) = (25- t 2) 3 where t ~ 0 is the time elapsed
in seconds. Find the time at which the acceleration of the object is zero. When is the object at
rest?

Answer 1

please help!

Answer 2

acc=0 means the double derivative=0
v=3(25-t^2)^2 (-2t)
a=3*2(25-t^2) (-2t) * 3(25-t^2)^2 (-2)=0
solve it

Answer 3

i now feel expanding it and solving it would have been easier

Answer 4

what does it mean when it says when the object is at rest?

Answer 5

the object it at rest means the velocity =0
for that put the v we got up =0

Answer 6

v=0 gives 25-t^2=0
hence t=5

Answer 7

gotcha!! thank you!!!!

Answer 8

i guess we get the same result for acc.

Answer 9

i'm still not getting it for some reason....for v(t) i have -6t(25-t^2)^2 is that correct?

Answer 10

yes

Answer 11

then you set that to 0? but how do you solve for t?

Answer 12

look now
-6t(25-t^2)^2=0
hence for it to be 0
either t=0 or 25-t^2=0
accordin to ur Question t is not 0 hence the second case

Answer 13

ohhh okay that makes sense now!!

Answer 14

and same for a(t)=24t(25-t^2) either t=0 or +/- 5? so you don't distribute 24t to what is inside the parenthesis you just set both to zero?