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MathCalculus

What Is Rate of Change? Instantaneous vs. Average Velocity for a Dropped Ball

The ball reaches the ground after 2 seconds. Its instantaneous velocity then is −64 ft/s, while its average velocity over the fall is −32 ft/s.

Question

A ball is dropped from a height of 64 feet. Its height above ground, in feet, tt seconds later is

s(t)=−16t2+64.s(t)=-16t^2+64.
  1. What is the instantaneous velocity of the ball when it hits the ground?
  2. What is the average velocity during its fall?

Answer

First find when the ball reaches the ground:

−16t2+64=0.-16t^2+64=0.

Thus

t2=4,t^2=4,

and because t≥0t\ge0, the ball hits the ground at

t=2 s.t=2\text{ s}.

1. Instantaneous velocity at impact

Instantaneous velocity is the derivative of position:

v(t)=s′(t)=−32t.v(t)=s'(t)=-32t.

At t=2t=2,

v(2)=−64 ft/s.\boxed{v(2)=-64\text{ ft/s}}.

The negative sign indicates motion in the downward direction if upward is taken as positive.

2. Average velocity during the fall

Average velocity over [0,2][0,2] is

s(2)−s(0)2−0=0−642=−32 ft/s.\frac{s(2)-s(0)}{2-0} =\frac{0-64}{2} =\boxed{-32\text{ ft/s}}.

The instantaneous rate describes the ball's velocity at one moment, while the average rate describes the net change in position per unit time over the entire interval.

Evidence boundary

The answer uses the position model s(t)=−16t2+64s(t)=-16t^2+64 exactly as given. The velocities describe this idealized model; no additional effects such as air resistance are introduced.

Sources

These references support the concepts and methods used in the explanation above.

What Is Rate of Change? Average and Instantaneous Velocity | Verla