Question

A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from rest to 20 m/s. What is the average power output of each? Average power over a time interval Δt is ΔE/Δt.

Answer 1

Answer:

$P_1 = 1166.7 Watt$

$P_2 = 2000 Watt$

Explanation:

Average power for the human sprinter is given as

$Power = \frac{\Delta E}{\Delta t}$

so we have

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}$

$P_1 = 1166.7 Watt$

Average power for greyhound is given as

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}$

$P_2 = 2000 Watt$