Velocity and displacement question?
An object is thrown straight up with a velocity, in ft/s, given by v(t)=−32t+65, where t is in seconds, from a height of 38 feet.
1) What is the object's maxiumum velocity?
2) What is the object's maximum displacement?
3) When is the object's displacement 0?
4) What is the object's maximum height?
- NCSLv 77 months agoFavorite Answer
1) max velocity at launch -- 65 ft/s
(agree with other post)
2) In order for the answers to parts 2 and 4 to be different, the launch point must be considered the location of zero displacement.
max s = v²/2g = 65²/64 = 66
3) s(t) = ∫ v(t) dt = -16t² + 65t + C
we're told s(0) = 38, so C = 38 and
s(t) = -16t² + 65t + 38
for displacement zero,
s(0) = 0 = -16t² + 65t + 38
which has a negative root we can ignore
and a positive root at
t ≈ 4.6
4) 38 + 66 = 104
Hope this helps!
- DylanLv 67 months ago
Strictly decreasing so v(0) is max
v(t) = 0, t = 65/32 s
v'(t) = s(t) = -16t^2 + 65t
Max at 65/32 s
s(0) = 38
Ans 2 + Ans 3