both planes were headed on the same route from alaska to hawaii. Plane #1 traveled at 520 mph and plane #2 traveled at 580 mph. If plane #1 took off one and a half hours before plane #2, how many hours did it take for plane #2 to catch plane #1?

Answer 1

Let's denote the time it takes for Plane #2 to catch up to Plane #1 as $t$ hours.

Since Plane #1 took off 1.5 hours before Plane #2, it has already been traveling for $t+1.5$ hours when Plane #2 starts.

We can calculate the distances traveled by each plane:

• The distance traveled by Plane #1 is: $${\text{Distance}}_{\text{Plane 1}}=\text{Speed}\times \text{Time}=520\text{ mph}\times (t+1.5)\text{ hours}$$

• The distance traveled by Plane #2 is: $${\text{Distance}}_{\text{Plane 2}}=\text{Speed}\times \text{Time}=580\text{ mph}\times t\text{ hours}$$

For Plane #2 to catch up to Plane #1, the distances must be equal: $$520(t+1.5)=580t$$

Now, we can solve for $t$:

1. Expand the left side: $$520t+780=580t$$

2. Rearrange the equation to isolate $t$: $$780=580t-520t$$ $$780=60t$$

3. Divide both sides by 60: $$t=\frac{780}{60}=13$$

Thus, it takes Plane #2 13 hours to catch up to Plane #1.