6 A long distance runner runs three laps a and in a desperate attempt to win she ri la10 mph. How much does this velo increase change her kinetic energy? A) doubles it B
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Answer:
A runner travels 1.5 laps around a circular track in a time of 50 s. The diameter of truck is 40 m and its circumference is 126 m. Find (a) the average speed of the runner and (b) the magnitude of the runner averwge's velocity. Be careful here: average speed depends on the distance traveled, whereas average depends on the displacement at the end of the particular journey.
a) The distance travelled by the runner (1.5 of the circle with the circumference 126 m) is:
l = 1.5\times 126m = 189ml=1.5×126m=189m
Since the time is t = 50st=50s, then the average speed is:
speed = {189m}{50s} = 3.78 m/sspeed=
t
l
a) The distance travelled by the runner (1.5 of the circle with the circumference 126 m) is:
l = 1.5\times 126m = 189ml=1.5×126m=189m
Since the time is t = 50st=50s, then the average speed is:
speed = \dfrac{l}{t} = \dfrac{189m}{50s} = 3.78 m/sspeed=
t
l
=
50s
189m
=3.78m/s
b) The displacement of the runner at the end of the jorney is equal to the diameter of the circle, since she completed 1.5 laps. The diameter is d = 40md=40m. Thus, the average velocity is:
velocity = {d}{t} = {40m}{50s} = 0.8m/svelocity=
t
d
=
50s
40m
=0.8m/s
Answer. a) 3.78 m/s, b) 0.8 m/s.
=
50s
189m
=3.78m/s
b) The displacement of the runner at the end of the jorney is equal to the diameter of the circle, since she completed 1.5 laps. The diameter is d = 40md=40m. Thus, the average velocity is:
velocity = \dfrac{d}{t} = \dfrac{40m}{50s} = 0.8m/svelocity=
t
d
=
50s
40m
=0.8m/s
Answer. a) 3.78 m/s, b) 0.8 m/s.