Speed of Truck Relative to Highway: Solving the Puzzle

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The discussion revolves around calculating the speed of a truck relative to the highway using a convex mirror. The car is traveling at 25 m/s, and the truck's image approaches the mirror at a speed that needs to be determined. The participants explore the relationship between the truck's speed and the image speed, noting that the image speed is not constant due to the nonlinear relationship between the distances involved. By applying the mirror equation and differentiating, they arrive at a solution indicating the truck's speed is approximately 51 m/s. The complexity arises from the changing speed of the image as it approaches the mirror.
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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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