I need help with all parts Careful measurements have been made of Olympic sprint
ID: 1493972 • Letter: I
Question
I need help with all parts
Careful measurements have been made of Olympic sprinters in the 100-meter dash. A simple but reasonably accurate model is that a sprinter accelerates at 3.6 m/s^2 for 31/3 s, then runs at constant velocity to the finish line. What is the race time for a sprinter who follows this model? Express your answer using two significant figures. A sprinter could run a faster race by accelerating faster at the beginning, thus reaching top speed sooner. If a sprinter's top speed is the same as in , what acceleration would he need to run the 100-meter dash in 9.5 s ? Express your answer using two significant figures. By what percent did the sprinter need to increase his acceleration in order to decrease his time by 5 % ? Express your answer using two significant figures.Explanation / Answer
distance travelled with acceleration x1 = ut + 0.5*a*t^2
u = initial velocity = 0
t = time interval = 10/3 s
a = 3.6
x1 = 0.5*3.6*(10/3)^2 = 20 m
speed of after t1 time
v = u + at = 0 + 3.6*10/3 = 12 m/s
time taken to the remaining distance t2 = (100-20)/12 = 20/3
total time taken T = 10/3 + 20/3 = 10 s <<-----------answer
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part B
T = 9.5 s
T = t1 + t2 ====> t2 = T - t1
top speed is same
v = a*t1
12 = a*t1
during t1 time
x1 = 0.5*a*t1^2 = 0.5*v*t1 = 6t1
x2 = v*t2 = v*(T-t1) = 12*(9.5-t1)
x1 + x2 = 100
100 = 6t1 + 12*(9.5-t1)
t1 = 7/3 s
acceleration a = v/t1 = 12/(7/3) = 5.14 m/s^2 <<<-------answer
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part C
% change = (5.14-3.6)/3.6 = 43 %
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