How to Solve the Distance Problem Using the Pythagorean Theorem

Understanding how to calculate the distance between two objects moving in perpendicular directions is a common problem in many fields, including engineering, physics, and even real-world navigation. Here, we will walk through a specific problem involving two cars using the Pythagorean theorem to determine when they will be 300 miles apart.

Understanding the Problem

Let's consider the scenario where two cars leave from the same point at 7:30 AM. One car travels east at 50 mph, and the other travels south at 60 mph. The question is, at what time will they be 300 miles apart?

Step-by-Step Strategy to Solve the Problem

The solution to this problem involves using the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Given

Car A: traveling east at 50 mph

Car B: traveling south at 60 mph

Required distance between the two cars: 300 miles

Step 1: Express Distance Traveled

Let ( t ) represent the time in hours after 7:30 AM. Then,

Distance traveled by Car A in ( t ) hours: ( 50t ) miles

Distance traveled by Car B in ( t ) hours: ( 60t ) miles

Step 2: Apply the Pythagorean Theorem

Since the cars are traveling in perpendicular directions, the distance between them forms the hypotenuse of a right-angled triangle. Therefore, we can write:

( (50t)^2 (60t)^2 300^2 )

Step 3: Solve for ( t )

Expanding and simplifying the equation:

( 2500t^2 3600t^2 90000 )

( 6100t^2 90000 )

( t^2 frac{90000}{6100} approx 14.7541 )

( t approx sqrt{14.7541} approx 3.85 ) hours

Step 4: Convert ( t ) to Hours and Minutes

( 0.85 ) hours ( 0.85 ) x ( 60 ) minutes approx 51 minutes

Thus, ( t approx 3 ) hours and ( 51 ) minutes.

Step 5: Determine the Time They Will Be 300 Miles Apart

Starting time: 7:30 AM

Adding 3 hours and 51 minutes:

7:30 AM 3:51 11:21 AM

Conclusion

The two cars will be 300 miles apart at approximately 11:21 AM. This technique can be applied to similar problems involving distances and perpendicular motion.

Additional Calculations

For alternative solutions, we can also verify the calculations:

If we assume the cars have traveled 5 miles and 6 miles respectively in ( x ) hours:

( sqrt{5^2 6^2} 300 )

( 25 36x^2 90000 )

( 36x^2 89975 )

( x^2 frac{89975}{36} approx 2499.3056 )

( x approx sqrt{2499.3056} approx 49.993 ) hours (approx 3 ) hours 50.5 minutes approximately

Hence, the two cars will be 300 miles apart at around 11:50.5 AM.

This confirms our original calculation.