Penny free fall

A man drops a penny from the top of a 500m tall building. After t seconds, the penny has fallen a distance of s meters, where s(t)=500-5t². Determine the average velocity between 1s and 5s.

Correct answer:

v =  30 m/s

Step-by-step explanation:

$$\begin{gathered} s(t)=500-5t^2 \\ {t}_1=1\text{s} \\ {t}_2=5\text{s} \\ v=\frac{ds}{dt} \\ v(t)=-5\cdot 2\cdot t^1=-10\cdot t \\ {v}_1=5\cdot 2\cdot t_1=5\cdot 2\cdot 1=10\text{m/s} \\ {v}_2=5\cdot 2\cdot t_2=5\cdot 2\cdot 5=50\text{m/s} \\ v=\frac{v_1+v_2}{2}=\frac{10+50}{2}=30\text{m/s} \end{gathered}$$

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