Answer:
1)0.325
2)
Explanation:
<u>Given:</u>
The angle that falling raindrops make with the vertical=
Let be the velocity of the raindrops and
be the velocity of the bus.
1)
2)Speed of the raindrops
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.
1 answer:
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
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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