Question

An object dropped from restfrom the top of a tall building falls y=16t^2 feet in the first t seconds.

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

Answer 1

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.