Question

Smallville and Big City are about 2800 miles apart. At 10:00 A.M an airplane leaves Smallville for Big City flying at 300 miles per hour. At the same time an airplane leaves Big City for Smallville flying at 400 miles per hour. How long will it be before the two airplanes meet?

Answer 1

Decide an origin. It could be either city (or anywhere for that matter), but let's just make Smallville the origin.

This means that Smallville is at position 0
This also means that Big City is at position 2800.

Answer 2

Then let the position of each plane be represented by \(x\) and \(y\), and let \(t\) be the number of hours passed.

At 10:00 A.M an airplane leaves Smallville for Big City flying at 300 miles per hour\[
x = 300t
\]
At the same time an airplane leaves Big City for Smallville flying at 400 miles per hour.
\[
y = 2800 - 400t
\]

All we have to do is set these equations equal to each other and solve for \(t\):\[
300t = 2800 - 400t
\]

Answer 3

This is based on the fact that if the distance of both planes from Smallville are the same, then they must be at the same position.

Answer 4

@truonghungngoc make sense?

Answer 5

Note that when they meet then this implies that the amount of time passed since each plane left each airport is the same. This must mean that:\[\bf \frac{ d_1 }{ t_1 }=300 \ and \ \frac{ d_2 }{ t_2 }=400, \ t_1=t_2\]\[\bf \implies \frac{d_1}{300}=\frac{d_2}{400}\]But we also know that \(\bf d_1+d_2=2800 \implies d_2=2800-d_1\). Making this substitution yields:\[\bf \implies \frac{ d_1 }{ 300 }=\frac{ 2800-d_1 }{ 400 } \implies 400d_1=840,000-300d_1\]Re-arranging yields:\[\bf 700d_1=840,000 \implies d_1=\frac{ 840,000 }{ 700 }=1,200 \ mi\]This must mean that \(\bf d_2=1,600 \ mi\). Now we just calculate how long it would've taken the first plane to travel \(\bf 1,200 \ mi\) (we could also calculate the time it took the other plane to travel 1,600 mi though both would yield the same answer since this is the distance each of the planes had travelled at the time they met). Since we know the plane with travel distance \(\bf d_1\) that it completed at the point of its meeting with the other plane, travels at 300 mi/hr then:\[\bf \frac{ 1,200 }{ t_1 }=300 \implies t_1=4 \ hours\]Hence it took \(\bf 4 \ hours \) for the two planes to meet. Do you understand?

@truonghungngoc

Answer 6

You didn't like my explanation?