Changing Rate Math Example 3

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Example 3

easy
A ball is thrown upward. Its height (m) is h(t)=−5t2+20th(t) = -5t^2 + 20t. Find the average rate of change from t=0t=0 to t=2t=2 seconds.

Solution

  1. 1
    h(0)=0h(0) = 0, h(2)=−5(4)+20(2)=−20+40=20h(2) = -5(4)+20(2) = -20+40 = 20 m.
  2. 2
    Average rate =h(2)−h(0)2−0=20−02=10= \frac{h(2)-h(0)}{2-0} = \frac{20-0}{2} = 10 m/s.

Answer

Average rate of change =10= 10 m/s
The average rate of change of height over time is the average velocity. For a projectile, this is positive on the way up, zero at the peak, and negative on the way down.

About Changing Rate

A changing rate of change means the output grows by different amounts for equal increases in input — the hallmark of nonlinear functions like quadratics and exponentials.

Learn more about Changing Rate →

More Changing Rate Examples

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Example 2 hard

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Example 4 medium

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