Problem

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Problem 14 Easy Difficulty

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.

Answer

a) $\begin{aligned} v_{1} &=6.00 \mathrm{m} / \mathrm{s} \\ v_{2} &=-1.71 \mathrm{m} / \mathrm{s} \\ v_{3} &=4.04 \mathrm{m} / \mathrm{s} \end{aligned}$
b) $v=3.49 \mathrm{m} / \mathrm{s}$

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Physics 101 Mechanics

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Chapter 2

Kinematics

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Video Transcript

so for party to find his average velocity, we could say the bar that would represent average velocity. This would equal Delta X over delta T, and then we can save the bars of one for the first interval. This would be delta acts of one divided by delta t someone. This is equaling 15.0 meters, divided by 2.50 seconds and this is equaling 6.0 meters per second. We can then say that for V bar sub to this would equal delta X up to divided by delta. T said too. This is equaling negative 3.0 meters divided by 1.75 seconds. And this is equally negative 1.71 meters per second for V bars of three. The third interval. This would be 21.0 meters, divided by 5.20 seconds and this is equaling 4.4 meters per second. These would be our three answers for part A for part B. Now they want us to find the average velocity throughout the entire time so we can say for part B, the average velocity would be equaling two Delta X of one plus Delta except two. We'll start to accept three divided by Dr Teeth of one plus Delta T substitute plus dotted T's of three, and this is gonna be equaling 2 33 0.0 meters, divided by 9.45 seconds. And so for party, the average velocity for the entire interval would be three point for nine meters per second. This would be our final answer for part B. That is the end of the solution. Thank you for what?

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