Question

#3b. Two bicyclists ride in the same direction. The first bicyclist rides at a speed of 8 mph.One hour later, the second bicyclist leaves and rides at a speed of 12 mph. How long will thesecond bicyclist have traveled when they catch up to the first bicyclist?I’m

Answer 1

2 hours

Step-by-step explanation:

`\text{Speed}=(Dis\tan ce)/(Time)`

The first bicyclist rides at a speed of 8 mph. Therefore:

`\begin{gathered} 8=(d)/(t) \\ \implies d=8t \end{gathered}`

One hour later, the second bicyclist leaves and rides at a speed of 12 mph.

Therefore, the time of the second bicyclist = (t-1) hours.

Therefore:

`\begin{gathered} 12=(d)/(t-1) \\ \implies d=12(t-1) \end{gathered}`

Since the second bicyclist will catch up to the first bicyclist, the distance traveled will be the same.

So:

`\begin{gathered} 8t=12(t-1) \\ 8t=12t-12 \\ 8t-12t=-12 \\ -4t=-12 \\ (-4t)/(-4)=(-12)/(-4) \\ t=3\text{ hours} \end{gathered}`

Therefore, the second bicyclist will have traveled for:

(t-1) = (3-1) =2 hours.