Math Challenges

Submissions for Problem #23

Problem #23

A ball is thrown into the air with a velocity of 40 ft/s. Its height in feet t seconds later is given by the equation below.

Find the average velocity of s over a time period beginning at t = 2 and lasting
a) 0.5 seconds
b) 0.1 seconds
c) 0.05 seconds
d) 0.01 seconds

Estimate the instantaneous velocity at t = 2

\[ s\left(t\right)=40t-16t^2 \]

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zapwai

Solution: \( \displaylines{m=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(t_0+\Delta t\right)-f\left(t_0\right)}{\Delta t}\\ \text{For us }t_0=2\text{ while }\Delta t\text{ varies.}f\left(2\right)=40\left(2\right)-16\left(2\right)^2=16.\\ a)\frac{f\left(2.5\right)-f\left(2\right)}{.5}=\frac{\left\lbrack40\left(2.5\right)-16\left(2.5\right)^2\right\rbrack-\left\lbrack16\right\rbrack}{.5}=-32\\ b)\frac{f\left(2.1\right)-f\left(2\right)}{.1}=\frac{\left\lbrack40\left(2.1\right)-16\left(2.1\right)^2\right\rbrack-\left\lbrack16\right\rbrack}{.1}=-25.6\\ c)\frac{f\left(2.05\right)-f\left(2\right)}{.05}=\frac{\left\lbrack40\left(2.05\right)-16\left(2.05\right)^2\right\rbrack-\left\lbrack16\right\rbrack}{.05}=-24.8\\ d)\frac{f\left(2.01\right)-f\left(2\right)}{.01}=\frac{\left\lbrack40\left(2.01\right)-16\left(2.01\right)^2\right\rbrack-\left\lbrack16\right\rbrack}{.01}=-24.16\\ \text{Estimate would be }-24} \)