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November 25, 2014

November 25, 2014

Posted by **cara** on Wednesday, September 2, 2009 at 2:09pm.

Find the speed of the object at t=3 seconds and confirmalgebraically. The answer is 96 feet/sec but how do i figure it out? cans omeone show me steps? thanks

- math -
**MathMate**, Wednesday, September 2, 2009 at 2:52pmThe given function is

y(t) = 16 t²

The speed is the derivative of y with respect to t, i.e.

y'(t)=dy/dt=16*(2t)=32t

So the speed at t=3 seconds,

y'(t)=y'(3)=32*3=96 ft/s.

To check algebraically, we take the distance travelled at 2.99 and 3.01 seconds, and divide the difference in distance by (3.01-2.99)=0.02 seconds.

y(2.99)=143.0416 ft

y(3.01)=144.9616 ft

approximate speed

= (y(3.01)-y(2.99))/(3.01-2.99)

= (144.9616-143.0416)/(0.02)

= 96 ft/sec. checks with previous calculations

Note: if you have not done calculus before, please let me know.

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