Blythe and Geoff compete in a 1.00-km race. Blythe's strategy is to run the first 600 m of the race at a constant speed of 4.10 m/s, and then accelerate to her maximum speed of 7.30 m/s, which takes her 1.00 min, and then finish the race at that speed. Geoff decides to accelerate to his maximum speed of 8.30 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3.00 min to reach his maximum speed. Assume all accelerations are constant. Required:
a. Calculate the time of Blythe's run.
b. Calculate the time of Geoff s run.

Explanation:

1.00 km = 1000 m .

Blythe's run :

Time taken to run 600 m speed

= distance / speed

T₁= 600 / 4.10 = 146.34 s

Next time T₂ = 1 min = 60 s

acceleration of Blythe

a = (7.30 - 4.10) / 60 = .053 m /s²

displacement during acceleration

= ut + 1/2 at²

= 4.10 x 60 + .5 x .053 x 60²

= 246 + 95.4

= 341.4 m

Rest of the distance to be covered = 1000 - ( 600 + 341.4 )

= 58.9 m

Time taken to cover this distance

T₃= 58.9 / 7.3 = 8.06 s

Total time = T₁ + T₂ + T₃ = 214.4 s

Geoff s run :

initial acceleration during first 3 min

= (8.3 - 0 ) / (3 x 60 )

= .046 m /s²

displacement

s = ut + 1/2 a t²

= 0 + .5 x .046 x ( 3 x 60 )²

= 745.2 m

Rest of the distance of race

= 1000 - 745.2 = 254.8 m

This distance is covered at speed of 8.3 m/s

time taken to cover this distance

T₂ = 254.8 / 8.3

= 30.7 s

Total time taken to complete the race

= 180 + 30.7

= 210.7 s .

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