Respuesta :
Answer:
-4
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.
Answer:
-4 mph
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
