Question

Two cars are traveling along perpendicular roads, car A at 40 mph, car B at 60 mph. At noon, when car A reaches the intersection, car B is 90 miles away and moving toward it. Find the rate at which the distance between the cars is changing at 1 P.M..

Answer 1

-4 mph

Step-by-step explanation:

Car A is traveling at 40mph.

Car B is traveling at 60mph

At noon, Car A reaches intersection while car B is 90 miles away and moving towards it.

Distance between car A and B is 90 miles.

da/dt = 40 mph

db/dt = 60mph

At 12 noon, car A is at (0,0) while car B is at (0,-90)

A is traveling along the x axis and B is traveling along the y axis.

At 1 pm, car A and B will be at (40,0) and (0,-30)

Car A has moved to the right along the x axis. Car B has moved up along the y axis by 60 because 1 hour passed since 12 pm

The rate at which the distance is changing is dd/dt

d = √(ax - bx) ^2 + (ay - by)^2

d = √(ax - 0)^2 + (0 - by) ^ 2

d = √ax^2 + by^2

d^2 = ax^2 + by^2

d^2 = a^2 + b^2

Differentiate implicitly

2d(dd/dt) = 2a(da/dt) + 2b(db/dt)

dd/dt = [a(da/dt) + b(db/dt)] / d

a = 40, b = -30

da/dt = 40

db/dt = 60

d^2 = 30^2 + 40^2

= 900 + 1600

= 2500

d = √2500

d = 50

= [40(40) + -30(60)]/ 50

dd/dt = (1600 - 1800) / 50

= -200/-50

dd/dt = -4 mph

Answer 2

-4

Step-by-step explanation:

Parameters given:

Speed of car A = 40 mph

Speed of car B = 60 mph

Initial position of car B relative to A = 90 miles

The attached diagram explains better.

For the sake of this solution, we assume that car A is moving on the y axis and car B is moving on the x axis.

First, calculate the position of both cars after 1 hour:

Car A:

y = 40 * 1 = 40 miles

Car B:

(60 * 1) = 60 miles.

Since car B is moving towards A and it was initially 90 miles, the position of B will be (90 - 60) miles

x = 30 miles

We can find the relative distance of both cars (s in the diagram) using Pythagoras theorem:

s² = x² + y²

s² = 30² + 40² = 2500

s = 50 miles

To find the rate at which this distance is changing with time, we differentiate the Pythagoras equation with respect to time:

s² = x² + y²

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

dx/dt = -60 mph (because car B is moving towards car A, hence it is approaching the -ve x axis)

dy/dt = 40 mph (it moves upward on the +ve y axis)

=> 2*50*ds/dt = (2*30*-60) + (2*40*40)

100*ds/dt = -3600 + 3200

100*ds/dt = -400

=> ds/dt = -400/100

ds/dt = -4

The negative sign indicates that the distance between both cars (s) is decreasing.