Question

the height of a ball (in meters) is modeled by $h(t)=5t^{2}+20t + 2$ find the average rate of change of height between $t = 1$ and $t = 3$.

Explanation:

Step1: Calculate \( h(1) \)

Substitute \( t = 1 \) into \( h(t)=5t^{2}+20t + 2 \)
\( h(1)=5\times(1)^{2}+20\times1 + 2=5 + 20+2=27 \)

Step2: Calculate \( h(3) \)

Substitute \( t = 3 \) into \( h(t)=5t^{2}+20t + 2 \)
\( h(3)=5\times(3)^{2}+20\times3 + 2=5\times9+60 + 2=45+60 + 2=107 \)

Step3: Use the average rate of change formula

The average rate of change formula is \( \frac{h(b)-h(a)}{b - a} \), where \( a = 1 \), \( b = 3 \)
\( \frac{h(3)-h(1)}{3 - 1}=\frac{107 - 27}{2}=\frac{80}{2}=40 \)
Wait, there is a mistake. The function should be \( h(t)=- 5t^{2}+20t + 2 \) (because the parabola opens down - ward, coefficient of \( t^{2}\) is negative).

Step1: Recalculate \( h(1) \) with \( h(t)=-5t^{2}+20t + 2 \)

Substitute \( t = 1 \)
\( h(1)=-5\times(1)^{2}+20\times1 + 2=-5 + 20+2=17 \)

Step2: Recalculate \( h(3) \) with \( h(t)=-5t^{2}+20t + 2 \)

Substitute \( t = 3 \)
\( h(3)=-5\times(3)^{2}+20\times3 + 2=-5\times9+60 + 2=-45+60 + 2=17 \)

Step3: Use the average rate of change formula again

\( \frac{h(3)-h(1)}{3 - 1}=\frac{17-17}{2}=0 \)

Answer:

\(0\)