How Fast is the Car Traveling Along the Highway?

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Homework Help Overview

The problem involves a highway patrol plane flying at a constant altitude and speed, detecting a car's distance decreasing over time. The context is related rates in calculus, focusing on the relationship between the speeds of the plane and the car.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the velocities of the car and the plane, questioning the interpretation of variables in the related rates equation. There is an exploration of how to express the car's velocity in relation to the plane's speed and the rate of distance decrease.

Discussion Status

Some participants are attempting to clarify the relationships between the variables and have provided expressions for the rates involved. There seems to be a productive exchange of ideas regarding the setup of the problem, although no consensus has been reached on the final velocity of the car.

Contextual Notes

Participants are working with specific values and relationships derived from the problem statement, but there is some confusion regarding the interpretation of certain variables and their rates. The discussion reflects an ongoing effort to reconcile these interpretations.

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Homework Statement


A highway patrol plane is flying 1 mile above a long, straight road, with constant ground speed of 120 m.p.h. Using radar, the pilot detects a car whose distance from the plane is 1.5 miles and decreasing at a rate of 136 m.p.h. How fast is the car traveling along the highway?


The Attempt at a Solution



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It's wrong. The value of x in the related rates equation is not an unknown. x^2+1=1.5^2. It's easy to find. And 'x' is the distance, not a velocity. dx/dT is not -120. It's a combination of the plane's velocity with the unknown velocity of the car. That's what you want to solve for.
 
Last edited:
So would you say that (velocity of car)\frac{dc}{dt} = 120 + \frac{dx}{dt}

so that

\frac{dh}{dt}=\frac{x}{\sqrt{x^2+1}}\frac{dx}{dt}

where

\frac{dx}{dt} = \frac{dc}{dt} - 120

\frac{dh}{dt}= -136

and

x=\sqrt{1.25}

so that \frac{dc}{dt} \approx -62.46 \approx 62.46 mph
 
Last edited:
e^(i Pi)+1=0 said:
So would you say that (velocity of car)\frac{dc}{dt} = 120 + \frac{dx}{dt}

so that

\frac{dh}{dt}=\frac{x}{\sqrt{x^2+1}}\frac{dx}{dt}

where

\frac{dx}{dt} = \frac{dc}{dt} - 120

\frac{dh}{dt}= -136

and

x=\sqrt{1.25}

so that \frac{dc}{dt} \approx -62.46 \approx 62.46 mph

Yes, I think that's more like it.
 

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