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.