both planes were headed from the outer boondocks to timbuktu. Plane number one traveled at 450 mph and plane number stwo at 500 mph. if plane number one took off one hour before plane two, how many hours did it take for plane number two to catch plane number one?

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 one hour earlier than Plane 2, by the time Plane 2 takes off, Plane 1 has already been flying for 1 hour. Thus, in that hour, Plane 1 travels:

\[ \text{Distance traveled by Plane 1 in 1 hour} = 450 \text{ miles per hour} \times 1 \text{ hour} = 450 \text{ miles} \]

Now, when Plane 2 starts flying, Plane 1 has a head start of 450 miles. After \( t \) hours of flying for Plane 2, the distance traveled by both planes can be expressed as follows:

• The distance traveled by Plane 1 in \( t + 1 \) hours is: \[ \text{Distance of Plane 1} = 450 \text{ mph} \times (t + 1) = 450(t + 1) \]

• The distance traveled by Plane 2 in \( t \) hours is: \[ \text{Distance of Plane 2} = 500 \text{ mph} \times t = 500t \]

To find when Plane 2 catches Plane 1, we set the distances equal to each other:

\[ 450(t + 1) = 500t \]

Now let's simplify and solve for \( t \):

\[ 450t + 450 = 500t \]

Subtract \( 450t \) from both sides:

\[ 450 = 500t - 450t \]

This simplifies to:

\[ 450 = 50t \]

Now, divide both sides by 50:

\[ t = \frac{450}{50} = 9 \]

So, it takes Plane 2 \( 9 \) hours to catch up to Plane 1.