Question

In: Physics

A sprinter accelerates at 3.08 m/s2 for a certain number of seconds during a 100.0 meter...

A sprinter accelerates at 3.08 m/s2 for a certain number of seconds during a 100.0 meter race. She runs the rest of the race at top speed and finishes the race in 10.59 s. What is her top speed?

Solutions

Expert Solution

Length of the race = L = 100 m

Total time taken by the sprinter to finish the race = T = 10.59 sec

Initial speed of the sprinter = V1 = 0 m/s

Acceleration of the sprinter = a = 3.08 m/s2

Top speed of the sprinter = V2

Time taken to reach the top speed = T1

V2 = V1 + aT1

V2 = 0 + aT1

V2 = aT1

Distance the sprinter travels while accelerating = L1

L1 = V1T1 + aT12/2

L1 = (0)T1 + aT12/2

L1 = aT12/2

Time period the sprinter travels at top speed = T2

T = T1 + T2

T2 = T - T1

Distance the sprinter travels at top speed = L2

L2 = V2T2

L2 = (aT1)(T - T1)

L2 = aTT1 - aT12

L = L1 + L2

L = aT12/2 + aTT1 - aT12

L = aTT1 - aT12/2

100 = (3.08)(10.59)T1 - (3.08)T12/2

100 = 32.62T1 - 1.54T12

1.54T12 - 32.62T1 + 100 = 0

T1 = 17.46 sec or 3.72 sec

T1 cannot be greater than the total time period which is 10.59 sec.

T1 = 3.72 sec

V2 = aT1

V2 = (3.08)(3.72)

V2 = 11.46 m/s

Top speed of the sprinter = 11.46 m/s


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