Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. **There will be no changes to other Yahoo properties or services, or your Yahoo account.** You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

# A stone is dropped off a bridge where the movement of the stone is given by h(t)= 90 -4.9t^2 where h is height in meters, t is the time ?

t is time in seconds.

a) determine the rock's average rate of change over the first 3 seconds.

b) Estimate the instantaneous rate of change at 3 seconds (test using an interval of 0.0001)

### 3 Answers

- az_lenderLv 72 months ago
(a) h(3) = 90 - (9)(4.9) = 45.9 m.

The ROCK does not HAVE an "average rate of change". Do they mean the average rate of change in the rock's position, or the average rate of change of the rock's speed? Your teacher is a MORON. (But not the only moron among teachers of Y!A students.)

Oh well. Let's say they want the average rate of change in the rock's position. That would be

-9(4.9)m / 3s = -14.7 m/s.

(b) Note that h(3.0001) = 90 - 4.9*(3.0001)^2

= 90 - 4.9*(3 + 0.0001)^2

= 90 - 4.9*3^2 - 4.9*2*3*0.0001 - 4.9*0.0001^2

= 90 - 44.1 - 4.9*0.0006 - a tiny, tiny term which we can ignore.

= 45.9 - 0.00294.

The change of -0.00294 occurred in 0.0001 second, so the instantaneous rate of change in its position near t = 3 is -29.4 m/s.

- MorningfoxLv 72 months ago
Do you mean the rate of change of the color? mass? cost? The problem doesn't say. I MIGHT assume it means the rock's height, but it could be speed, or acceleration?

- rotchmLv 72 months ago
a) avg rate of change is Δh / Δt. Evaluating, you'll get?

b) Again, Δh / Δt for the times t=3 and t = 3.0001. You get?

Show your answers here and we will tell you if you got it okay or not.

Don't forget to vote me best answer for being the first of correctly walk you through without spoiling out the answers. That way gives you a chance to work at it and to get good at it!