Answer:
a) 30.726m/s and b) 5.5549s
Step-by-step explanation:
a.) What was Chris Huber’s speed in meters per second(m/s)?
Given the distance and time, the formula to obtain the speed is
.
Applying this to our problem we have that
.
So, Chris Huber’s speed in meters per second(m/s) was 30.726m/s.
b) What was Whittingham’s time through the 200 m?
In a) we stated that
. This formula implies that
.
First, observer that
.
Then, Sam Whittingham speed was equal to Chris Huber’s speed plus 5.2777 m/s. So, ![v=30.726\frac{m}{s} +5.2777\frac{m}{s}= 36.003 m/s.](https://tex.z-dn.net/?f=v%3D30.726%5Cfrac%7Bm%7D%7Bs%7D%20%2B5.2777%5Cfrac%7Bm%7D%7Bs%7D%3D%2036.003%20m%2Fs.)
Then, applying 1) we have that
![t=\frac{200m}{36.003m/s}=5.5549s.](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B200m%7D%7B36.003m%2Fs%7D%3D5.5549s.)
So, Sam Whittingham’s time through the 200 m was 5.5549s.