Question

Two cars are approaching an intersection on roads that are perpendicular to each other. Car A is north of the intersection and traveling south at 60 mph. Car B is east of the intersection and traveling west at 45 mph. How fast is the distance between the cars changing when car A is 14 miles from the intersection and car B is 48 miles from the intersection? (Round your answer to two decimal places.)
mph?

Answer 1

Solution :-

from given data we have :-

• dx/dt = (-60) { as speed of A is 60mph towards intersection we took negative value.)
• dy/dt = (-45).
• Let original distance between them is = s miles.
• x = 14 miles .(Distance from intersection)
• y = 48 miles.
• ds/dt = ? ( Distance between them changing at which rate.)

using pythagoras theorem between x and y, we get,

→ s² = x² + y²

→ s² = (14)² + (48)²

→ s² = 196 + 2304

→ s² = 2500

square - root both sides,

→ s = 50 miles.

Now,

→ s² = x² + y²

Differentiate with respect to t , we get,

→ 2 * s * ds/dt = 2 * x * dx/dt + 2 * y * dy/dt

putting all values we get,

→ 2 * 50 * ds/dt = 2 * 14 * (-60) + 2 * 48 * (-45)

→ 100 * ds/dt = - 1680 - 4320

→ 100 * ds/dt = - 6000

dividing both sides by 100,

→ ds/dt = (-60) (Ans.) { Negative answer means that the distance is decreasing.}

Hence, the distance between them is decreasing at a rate of 60 mph.