Q23PE - College Physics (Urone)

Question

If a marathon runner averages\({\bf{9}}.{\bf{5}}{\rm{ }}{{{\rm{mi}}} \mathord{\left/ {\vphantom {{{\rm{mi}}} {\rm{h}}}} \right. \\} {\rm{h}}}\), how long does it take him or her to run a \(26.22{\rm{ mi}}\) marathon?

Step-by-Step Solution

Answer

Time required by him or her to cover \(26.22{\rm{ mi}}\) will be \(2.76{\rm{ h}}\).

Step 1: Given data

Consider the given data as below.

Average speed, \(v = 9.5{\rm{ }}{{{\rm{mi}}} \mathord{\left/ {\vphantom {{{\rm{mi}}} {\rm{h}}}} \right. \\} {\rm{h}}}\)

Distance, \(d = 26.22{\rm{ mi}}\)

Step 2: Required formula:

If you know how far something has traveled and how long it took to get there, 

you can calculate its average speed. 

The speed at which an object's location changes in any direction.

 The distance traveled in relation to the time it took to travel that distance is 

how speed is defined.

\(v = \frac{d}{t}\)                                                                                                                    ….. (1)

Here, \(v\) is the velocity, \(d\) is the distance, and \(t\) is time.

 

Step 3: Determine the time:

The time takes for him or her to run amarathon is define by rearranging equation (1).

\(t = \frac{d}{v}\)

Substitute \(9.5{\rm{ }}{{{\rm{mi}}} \mathord{\left/ {\vphantom {{{\rm{mi}}} {\rm{h}}}} \right.} {\rm{h}}}\) for \(v\) and \(26.22{\rm{ mi}}\) for \(d\) in the above equation.

\(\begin{array}{c}t = \frac{{26.22{\rm{ mi}}}}{{9.5{\rm{ }}{{{\rm{mi}}} \mathord{\left/ {\vphantom {{{\rm{mi}}} {\rm{h}}}} \right.} {\rm{h}}}}}\\ = 2.76{\rm{ h}}\end{array}\)

Step 4: Conclusion:

Time required by him or her to cover \(26.22{\rm{ mi}}\) will be \(2.76{\rm{ h}}\).