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