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# 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

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• (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.

• 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?

• 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?

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