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Semenov [28]
3 years ago
7

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.

Physics
1 answer:
My name is Ann [436]3 years ago
3 0
Define
t = time (s),
dt = change in time (s)
 d =distance (m).

The scenario is as follows:
t = 0 s: Begin. Quarterback runs forward.
d = 0

t = 2.5 s: Quarterback is tackled and pushed backward.
d = 15 m

t = 2.5+1.75 = 4.25 s: Quarterback resumes the run forward.
d = 15 - 3 = 12 m

t = 4.25+5.2 = 9.45 s: Quarerback runs another 21 m forward.
d = 12+21 = 33 m

The graph showing the scenario is shown below.

Calculations:
Interval 1:
Average velocty = 15/2.5 = 6.0 m/s

Interval 2:
Average velocity = (12 - 15)/(1.75) = - 1.7 m/s

Interval 3:
Average velocity = (33 - 12)/5.2 = 4.0 m/s

Entire motion:
Average velocity = (33 - 0)/9.45 = 3.5 m/s

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The time when the two players will collide is 0.96 s.

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The time when the two players collide, their displacement is equal or the difference in their position will be zero.

X_1 - X_2 = 0\\\\0.1 - 3.9 t - (-6.3 + 2.8t) = 0\\\\0.1 -3.9t + 6.3 -2.8t = 0\\\\6.4 -6.7t = 0\\\\6.7t = 6.4\\\\t = \frac{6.4}{6.7} = 0.96 \ s

Thus, the time when the two players will collide is 0.96 s.

Learn more here: brainly.com/question/18033352

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