One more.

A rock thrown vertically upward from the surface of the moon at a velocity of 32 meters per second reaches a height of s(t)=32t-0.8t^2 meters in t seconds.

Part A: Find the rock's velocity and acceleration as functions of time.

Part B: How long did it take the rock to reach its highest point?

I did part A and got v(t)= 32-1.6t and a(t)=-1.6 but i'm not sure what to do for part B.

Question

One more.

A rock thrown vertically upward from the surface of the moon at a velocity of 32 meters per second reaches a height of s(t)=32t-0.8t^2 meters in t seconds.

Part A: Find the rock's velocity and acceleration as functions of time.

Part B: How long did it take the rock to reach its highest point?

I did part A and got v(t)= 32-1.6t and a(t)=-1.6  but i'm not sure what to do for part B.

Vocaloid · Answer

at the highest point, the velocity is 0, so you can simply set the velocity equation equal to 0 and solve for t

mxddi3 · Answer

So would highest point be just considered at rest then? And we'd get t=20 seconds for the answer?

and overall, we're taking the derivative of the given equation to find the velocity, and then taking the derivative of the velocity to find acceleration correct?

Vocaloid · Answer

yes to both parts