Percent kinematics

A runner completes a 12km race with an average velocity ( v t o t a l \overline { { v }_{ total } } ) of 4 m/s 4\text{ m/s} . She does this in a certain amount of time, and after running for the first 16% of this time, her average velocity is 8 m/s 8\text{ m/s} when she reaches a checkpoint. Afterwards, she continues running for the same amount of time again. By the end of this time, she covers the first 50% of the race's distance.

Using the information given, calculate the runner's average velocity, v m i d \overline { { v }_{ mid } } , from the checkpoint up to this halfway point. Express your answer as the percent decrease from v m i d \overline { { v }_{ mid } } to v t o t a l \overline { { v }_{ total } } .

Details and Assumptions

Do not calculate the average velocity as time-weighted; use the following formula: v = Δ x Δ t = x f x i t f t i \overline { v } =\frac { \Delta x }{ \Delta t } =\frac { { x }_{ f }-{ x }_{ i } }{ { t }_{ f }-{ t }_{ i } } Treat all acceleration as constant; there is no need for calculus.


Picture credits: F5 St. Patrick’s Day Dash by Erika Schultz, The Seattle Times
36 % 36\% 11. 1 % 11.\overline { 1 }\% 7. 7 % 7.\overline { 7 }\% 33. 3 % 33.\overline { 3 }\% 25 % 25\%

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1 solution

Zee Ell
Jul 20, 2016

Total time = 12000 m 4 m / s = 3000 s \text {Total time} = \frac {12000 \ m}{4 \ m/s} = 3000 \ s

Time and distance between the start and the checkpoint:

3000 s × 16% = 480 s

480 s × 8 m/s = 3840 m

Distance and average velocity between the checkpoint and the halfway point:

12000 m × 0.5 - 3840 m = 2160 m

2160 m 480 s = 4.5 m / s \frac {2160 \ m} {480 \ s} = 4.5 \ m/s

Percentage decrease:

1 4 m / s 4.5 m / s = 0. 1 ˙ = 11. 1 ˙ % 1- \frac {4 \ m/s}{4.5 \ m/s}= 0. \dot{1} = \boxed {11. \dot {1} \%}

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