Question

An athlete cycles 8 mi/hr faster than the run. Every morning they cycle 4 miles and run 2.5 miles for a total of one hour of exercise. How fast do they run?

Answer 1

Let the speed for running be x

Let the speed for cycling be y

Given that the athlete cycles 8 miles/hr faster than he runs, we have

`y\text{ = (x +8) ------ equation 1}`

Every morning, the athlete cycles 4 miles and runs 2.5 miles for a total of 1 hour.

This implies that the total time for cycling and running is 1 hour.

`\begin{gathered} \text{speed}=\frac{\text{distance}}{\text{time}} \\ \Rightarrow time=(dis\tan ce)/(speed) \end{gathered}`

Time for cycling:

`t_{\text{cycling}}=\frac{4\text{miles}}{x\text{ miles/hr}}`

Time for running:

`t_(cycling)=\frac{2.5\text{ miles}}{y\text{ miles/hr}}`
`\begin{gathered} t_(cycling)+t_(running)=1\text{ hour} \\ (4)/(x)+(2.5)/(y)=1-------\text{ equation 2} \end{gathered}`

Substitute equation 1 into equation 2

solving the quadratic equation by factorization, we have

`\begin{gathered} x^2+1.5x-32=0 \\ (x-4.956)(x+6.456)=0 \\ x-4.956=0 \\ x\Rightarrow4.956 \\ x+6.456=0 \\ \Rightarrow x=-6.456 \end{gathered}`

since x and y are the speeds of running and cycling, their values cannot be negative,

Thus, x= 4.956 miles/hr

Hence, they run at 4.956 miles/hr