Speed of Truck Relative to Highway: Solving the Puzzle

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Homework Help Overview

The problem involves determining the speed of a truck relative to a highway while considering the effects of a convex mirror. The scenario includes a car moving at a constant speed and the truck approaching the mirror, with the image of the truck changing position as it nears the mirror.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the speeds of the truck and its image, questioning the assumption of constant speed. Some explore the implications of differentiating the mirror equation to understand how the image speed varies.

Discussion Status

Participants are actively engaging with the problem, attempting to apply calculus to derive relationships between the variables. There is recognition of the non-constant nature of the image speed, and some have proposed methods to analyze the situation further.

Contextual Notes

There is an acknowledgment of potential misunderstandings regarding the speed of the image and the need for clarity on the relationship between the distances involved. The discussion reflects a mix of confidence and uncertainty among participants as they navigate the problem.

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Homework Statement



You are in your car driving on a highway at 25 m/s when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of when the truck is 2.0 m from the mirror, what is the speed of the truck relative to the highway?

Homework Equations


The Attempt at a Solution



(sorry my bad english)
Assuming both the truck and the car speeds to be constant, the truck takes the same time interval as its image to reach the vertex of the mirror. So,

1 / f = 1 / s + 1 / s'
-1 / 0.75 m = 1 / 2 m + 1 / s'
s' = -0.54 m
t = -0.54 m / -1.9 m /s
[ (V - 25) m / s ] t = 2 m
but solving this for V I didnt find the correct value: 51 m / s. Maybe the speed is'nt constant?

My sketch:

1264etw.jpg
 
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The speed of the image will not be constant. You can see this by differentiating the 1/f formula.
 
I don't know how to do this :frown:
 
kent davidge said:
I don't know how to do this :frown:
The derivative of s with respect to time, ds/dt, is the velocity, v. What is the derivative of 1/s? Use the chain rule.
 
hmmmm let me try...

1 / f = 1 / V dt + 1 / V' dt
1 / f = 1 / dt (1 / V + 1 / V')

but how can I solve this derivative? would I need to express f as function of t?
 
Last edited:
kent davidge said:
hmmmm let me try...

1 / f = 1 / V dt + 1 / V' dt
1 / f = 1 / dt (1 / V + 1 / V')

but how can I solve this derivative? would I need to express f as function of t?
f is a constant. What is the derivative of a constant?
The chain rule says d/dt(1/s)= (ds/dt)(d/ds(1/s)). What is d/ds(1/s)?
 
ohh yah

(d / dt) 1 / f = 0
0 = - [(V - 25 m / s) / s²] - V' / s'²
V ≅ 51 m / s

Thanks :smile:

but why the image speed is not constant?
 
Last edited:
kent davidge said:
ohh yah

(d / dt) 1 / f = 0
0 = - [(V - 25 m / s) / s²] - V' / s'²
V ≅ 51 m / s

Thanks :smile:

but why the image speed is not constant?
Because the relationship between the two distances is nonlinear.
 

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