Respuesta :
Answer:
The rate of change of the distance between the cars at that instant is 5.46 meters per second
Step-by-step explanation:
Speed of first car = 7 m/s
Speed of second car = 3 m/s
At instant, distance of first car from intersection = 5 meters
distance of second car from intersection = 12 meters
Therefore, we have
Distance between both cars at intant = √(12² + 5²) = 13 m
Rate of cahnge of distance is given by;
2ddd/dt = d(x² + y²)/dt = 2xdx/dt + 2ydy/dt
= 26dd/dt = 2×5×7 + 2×12×3 = 142
dd/dt = 142/26 = 5.46 m/s.
That is the rate of change of the distance between the cars at that instant = 5.46 meters per second.
The rate of change of the distance between the cars at that instant is 5.46 meters per second.
Given that,
Two cars are driving away from an intersection in perpendicular directions.
The first car's velocity is 7 meters per second and the second car's velocity is 3 meters per second.
At a certain instant, the first car is 5 meters from the intersection and the second car is 12 meters from the intersection.
We have to find,
What is the rate of change of the distance between the cars at that instant (in meters per second)?
According to the question,
Let, the distance traveled by the first car is x,
And the distance traveled by the second car is y,
and z is the distance between the cars at that instant.
Two cars are driving away from an intersection in perpendicular directions.
Then,
The distance traveled by both cars is,
[tex]\\\rm The \ distance \ traveled \ by \ both \ car =\sqrt{ x^2+y^2}\\\\z^2 = { x^2+y^2}\\\\z = \sqrt{ x^2+y^2}\\[/tex]
At a certain instant, the first car is 5 meters from the intersection and the second car is 12 meters from the intersection.
Then,
The distance traveled by both the cars from the point of intersection is,
[tex]\rm The \ distance \ traveled \ by \ both \ car =\sqrt{5^2+12^2}\\\\\rm The \ distance \ traveled \ by \ both \ car =\sqrt{25+144}\\\\\rm The \ distance \ traveled \ by \ both \ car =\sqrt{169}\\\\\rm The \ distance \ traveled \ by \ both \ car =13 \ meter[/tex]
To find the rate of change of the distance between the cars at that instant differentiate the function with respect to x,
[tex]\dfrac{d(z^2)}{dx} = \dfrac{d(x^2)}{dx} + \dfrac{d(y^2)}{dx}\\\\2\dfrac{dz}{dx} = 2x \dfrac{dx}{dt}+ 2\dfrac{dy}{dx}\\\\\\\\[/tex]
The first car's velocity is 7 meters per second and the second car's velocity is 3 meters per second.
Substitute all the values in the equation,
[tex]2z\dfrac{dz}{dx} = 2x \dfrac{dx}{dt}+ 2\dfrac{dy}{dx}\\\\ z\dfrac{dz}{dx} = x \dfrac{dx}{dt}+ y\dfrac{dy}{dx } \\\\ 13\dfrac{dz}{dx} = 5 \times 7+ \ 3 \times 12 \\\\ 13 \dfrac{dz}{dx} = 35+ 36 \\\\ 13\dfrac{dz}{dx} = 71 \\\\ \dfrac{dz}{dx} = \dfrac{71}{13}\\\\ \dfrac{dz}{dx} = 5.46[/tex]
Hence, the rate of change of the distance between the cars at that instant is 5.46 meters per second.
For more details refer to the link given below.
https://brainly.com/question/9351049
