To solve this problem, we need to determine the rate at which the distance between the two cars is increasing at the moment when the first car is 120 miles from the junction.
Let's break this down step-by-step:
1. Understand the problem setup:
- The first car is traveling east at 40 mph.
- The second car is traveling south at 30 mph.
- We need to find the rate at which the distance between the two cars is increasing when the first car has traveled 120 miles east.
2. Determine the time it takes for the first car to travel 120 miles:
- Rate of the first car (east) = 40 mph
- Distance traveled by the first car (east) = 120 miles
- Time taken [tex]\( t \)[/tex] = Distance / Rate = 120 miles / 40 mph = 3 hours
3. Calculate the distance the second car has traveled:
- Rate of the second car (south) = 30 mph
- Time traveled by the second car = 3 hours
- Distance traveled by the second car (south) = Rate Time = 30 mph 3 hours = 90 miles
4. Use the Pythagorean theorem to find the distance between the two cars:
- Let [tex]\( x \)[/tex] be the distance traveled by the first car (120 miles).
- Let [tex]\( y \)[/tex] be the distance traveled by the second car (90 miles).
- The distance between the two cars ([tex]\( z \)[/tex]) is the hypotenuse of a right triangle with legs [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
- Therefore, [tex]\( z = \sqrt{x^2 + y^2} = \sqrt{(120)^2 + (90)^2} = \sqrt{14400 + 8100} = \sqrt{22500} = 150 \)[/tex] miles.
5. Determine the rate at which the distance between the two cars is increasing:
- Let [tex]\( z \)[/tex] be the distance between the two cars.
- We want to find [tex]\( \frac{dz}{dt} \)[/tex] when the first car is 120 miles east.
- According to the Pythagorean theorem, [tex]\( x^2 + y^2 = z^2 \)[/tex].
- Differentiate both sides with respect to time [tex]\( t \)[/tex]: [tex]\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \)[/tex].
- Simplify to: [tex]\( x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt} \)[/tex].
- Plugging in the values: [tex]\( x = 120 \)[/tex] miles, [tex]\( \frac{dx}{dt} = 40 \)[/tex] mph, [tex]\( y = 90 \)[/tex] miles, and [tex]\( \frac{dy}{dt} = 30 \)[/tex] mph.
6. Solve for [tex]\( \frac{dz}{dt} \)[/tex]:
- [tex]\( 120 \cdot 40 + 90 \cdot 30 = 150 \cdot \frac{dz}{dt} \)[/tex]
- [tex]\( 4800 + 2700 = 150 \cdot \frac{dz}{dt} \)[/tex]
- [tex]\( 7500 = 150 \cdot \frac{dz}{dt} \)[/tex]
- [tex]\( \frac{dz}{dt} = \frac{7500}{150} = 50 \)[/tex] mph
Therefore, the rate at which the distance between the two cars is increasing when the first car is 120 miles from the junction is 50 mph.
