Question

The Kentucky Derby, the first leg of horse racing’s Triple Crown, was won by a time of 122.2 s. If the race covers 2011.25 m, what was the average speed in a) m/s and b) in mi/h?
20 Extra points! Please help

Answer 1

The answer is 789.25 which you’d subtract 2011.25-122.2 I think sry if I’m wrong

Answer 2

Answer:

a) $16.45 \frac{m}{s}$ b) $36.94 \frac{mi}{h}$

Explanation:

The average speed is given by the formula: $\frac{distance}{time}$

The distance covered is 2011.25m , and the time is 122.2s.

Replacing in the formula: $speed_{avg} = \frac{2011.25 m}{122.2s}$

Resulting in a) $speed_{avg} = 16.45 \frac{m}{s}$.

To get the result in mi/h, do the following conversion:

Knowing : 1 mi = 1603 m   and  1 h = 3600 seg

• $16.45\frac{ m}{s}*(\frac{1mi}{1603m } ) *(\frac{3600s}{1h} ) = 36.94\frac{mi}{h}$