# Second Half of a Trip - Speeding tickets

kchurchi
Δ

## Homework Statement

In many countries, automatic number plate recognition is used to catch speeders. The system takes a time-stamped license plate photo at one location (like an on-ramp to a freeway), and then takes a second time-stamped photo at a second location a known distance from the first location (like an exit ramp off a freeway). Since the systems knows both the distance traveled and the time you took to travel that distance, it can determine if, on average, you were speeding, and subsequently mail the car owner a complementary ticket.

Let us say you are traveling along such a highway where the speed limit is 90 km/h (25m/s). Watching the distance markers on the side of the road, you discover at the half-way point you have been traveling at a speed of 114 km/h (31.67 m/s). What maximum constant speed must you travel for the last half of the trip to avoid your parents (who own the car) getting a speeding ticket in the mail?

## Homework Equations

v(avg) = (v(max)-v(initial))/Δt
v(avg) = v(speed limit)

## The Attempt at a Solution

I set the two equations equal to one another to find v(max) since we are given v(initial) and given the speed limit. However, I don't know how to find Δt. I have tried several different ways on paper and come to no conclusion. Help please.

Mentor
Average velocity depends upon the total distance and total time, not (at least not directly) on the two velocities involved unless the times for each travel segment happen to be equal.

Think about how you might write (symbolic) expressions for the travel times of each segment in terms of the segment distances and speeds. Think about an expression for the average speed given the two segment times and total distance.

kchurchi
So you are saying that the times for both journeys should be equal. This makes sense.
Let's say the time travelled at 31.67m/s is t(1) and the time travelled at v(max) is t(2).

t(1) + t(2) = t where t is time if he had travelled at the speed limit the whole way.

Let's also say that x equals the full distance of the highway.

(1/2)*x is the distance travelled in t(1) and (1/2)*x is the distance travelled in t(2).

v(initial) = Δx/Δt assuming that the initial values for both x and t equal zero we can say

v(initial) = ((1/2)*x)/t(1)

Similarly...

v(max) = ((1/2)x)/t(2) → v(max)*t(2) = (1/2)*x

Therefore...

v(initial) = (v(max)*t(2))/t(1)

v(max) = (v(initial)*t(1))/t(2)

but I still don't know time. Am I going in circles?

Mentor
Well, at least the circles are getting smaller... Write expressions (in terms of speed and distance) for the two times, t1 and t2. The total time must be t1 + t2, right? So in an expression for the average velocity, t1 and t2 can be replaced with these expressions... you should find that the irrelevant "unknowns" cancel.

kchurchi
:)

Okay so instead, I could write this...

v(max) = ((1/2)*x)/t(2) → t(2) = ((1/2)*x)/v(max)

v(initial) = ((1/2)*x)/t(1) → t(1) = ((1/2)*x)/v(initial)

Thus...

t(2) + t(1) = ((1/2)*x)/v(max) + ((1/2)*x)/v(initial) = t

Alright, I am with you up to here. Now suppose I do this...

v(speed limit) = x/t where x is the total distance and t is the total time

t = x/v(speed limit)

Therefore I can set this expression equal to the one above...

((1/2)*x)/v(max) + ((1/2)*x)/v(initial) = x/v(speed limit)

(1/2)*x*[1/v(max) + 1/v(initial)] = x/v(speed limit)

x cancels out and I mutltiply both sides by 2...

1/v(max) + 1/v(initial) = 2/v(speed limit)

1/v(max) = 2/v(speed limit) - 1/v(initial)

I take the inverse of both sides and...

v(max) = [2/v(speed limit) - 1/v(initial)]^(-1)

I'll give that a try!

kchurchi
Wow thanks for the help!! Got the right answer :D

Nogueira
you could also do a much simpler way. Since half the way was made at 31.67m/s, if the other half was made at 25m/s, the simple average would have been 28.335m/s, which is 3.335 m/s higher than the average required., now, if you double the value (because the average is made by dividing by 2 the two values), and subtract 25m/s, the given value is 18.33 m/s which is the correct answer