Collecting different answers to a simple problem

Question

I am collecting different answers to the problem below for possible publication in a pedagogical note I am writing. Please post freely (avoiding repetitions), and let your imagination go wild with alternative methods of solution. If I decide to quote one of your answers I will contact you for permission.

Edited for clarity:

While driving on a loooong straight highway with no other traffic, your GPS shows that your current speed is $V+\varepsilon$, where $V$ is the official speed limit in MPH. this speed will remain constant in the problem. The GPS also displays an estimated time of arrival (ETA) to your destination which is $d/V$, where $d$ is the remaining distance (in practice, a real GPS usually adds corrections for traffic, stoplights, etc.) Notice that $d$ is unknown to you.

Since you are driving faster than the speed used for the GPS estimate, this estimate will go down. Find a formula in terms of $V$ and $\varepsilon$ for the time (in minutes) that is needed to reduce the ETA by one minute.

I have read my post a couple of times and do not find it ambiguous. Christian Blatter below go it right. I am asking for alternative ways to reach the same solution. — Rodrigo A. Pérez, Aug 23 '13 at 01:44
A pedagogical note? About how to formulate math questions? >"While driving on a looong straight highway with no other traffic, your GPS displays your current speed $V+ε$, where $V$ is the official speed limit in MPH." The first thing that came to my mind was this: If I drive in a car and my display says "$V+ε$" I probably panic and die. — Nikolaj-K, Aug 22 '13 at 12:49
You said the current speed is V+eps, so shouldn't the ETA estimate already have changed? You should lay down the full assumption about ETA computation: The driver is currently driving at speed V. The ETA is what you expect. Suppose the driver instantaneously increases the speed to V+eps. How long until the ETA (which also changed instantaneously) reduces by a minute compared to the start of the problem? In fact, better to word the question without ETA at all! — Evan, Aug 25 '13 at 15:31
@RodrigoA.Pérez Ah now it's clear. So the GPS doesn't use your current speed in the ETA estimate. — Evan, Aug 25 '13 at 17:24
@RodrigoA.Pérez Hmm.. Think there is an additional confusion: You seem to want the moment the ETA of the GPS will display a minute less, and I provided that (unless I am missing something silly). But you said Christian Blatter's solution is correct? — Evan, Aug 26 '13 at 01:25

score 3 · Answer 1 · answered Aug 22 '13 at 08:22

3

I interpret the problem as follows: Instead of driving at speed $V$ you drive at speed $V+\epsilon$ for a certain time span $t$ in order to make good a time amount $\Delta$ (one minute) on the ETA.

During this time span $t$ a certain distance $s$ is traveled. When driving at speed $V$ you would need the time amount $t+\Delta$ to travel this distance. Therefore we have the following equation: $$(V+\epsilon)t=s=V(t+\Delta)\ .$$ Solving for $t$ we obtain $$t={V\over\epsilon}\Delta\ .$$ Here $t$ and $\Delta$ are measured in minutes, while $V$ and $\epsilon$ are measured in miles per hour.

answered Aug 22 '13 at 08:22

Christian Blatter

226,825

Thanks. This is straightforward. Now I'm hoping to see other solutions that have some 'twist' to how the setting is thought about. – Rodrigo A. Pérez Aug 22 '13 at 12:50
This is probably the intended interpretation, but then the original question should reflect that. I was trying to find this kind of interpretation that gives a simple answer, so this is nice! – Evan Aug 22 '13 at 14:39

score 1 · Answer 2 · answered Aug 25 '13 at 20:20

Since you're collecting solutions, here's a different way to look at this problem. Since you want to arrive at a time $\Delta$ sooner than the GPS thinks you will, you can pretend you had a head start of $\Delta$ and traveled at velocity $V$. In reality, you started without that lead and traveled at a rate of $V+\epsilon$. Graph the actual and imagined positions together and you're looking for where the two lines cross (i.e., when the distances are the same). Doing so, you'll get this picture enter image description here

where the red line is the actual situation and the blue line is the imagined situation. Of course, the intersection time is exactly what Christian found, $t = V\Delta/\epsilon$.

ok so we still have two different interpretations for the what the question asks . What you and Christian have at the end is the actual time traveled to the destination given the information in the problem. I have the time it takes for the GPS estimates, which for some reason only uses the speed limit of the road, to drop by 1 minute. so maybe the OP can clarify once again what is desired... — Evan, Aug 25 '13 at 22:05

score 1 · Answer 3 · answered Aug 25 '13 at 20:45

We want to reach our destination $\delta$ units earlier in time. Let $t$ denote the time period for which we should travel at speed $V+\epsilon$. This means that we can travel an extra distance of $t\cdot\epsilon$ before reaching our destination. If we travel the distance $V\delta$ beforehand, we'll reach our destination $\delta$ units of time earlier. Thus, $t\epsilon=V\delta$.

Evan · Answer 4 · 2013-08-25T17:43:15.920

0

So the ETA given the distance remaining $d$ will show $d/V*60$ minutes where $d$ is unknown to us.

Given time $t$ (in minutes), we reduce the distance by $(V+\epsilon)t/60$.

The ETA at time $t$ would show $\frac{d - (V+\epsilon)t/60}{V}*60$.

Comparing the two ETA's, the difference is $(V+\epsilon)t/V$ which setting to $1$ minute, shows that $t = \frac{V}{V+\epsilon}$ (in minutes).

edited Aug 25 '13 at 17:43

answered Aug 22 '13 at 07:02

Evan

3,861

score 0 · Answer 5 · answered Aug 22 '13 at 13:35

When ETA=T speed should is $V+\epsilon$, So with ETA=$T-1$ speed should be $\Large \frac{(T-1)(V+\epsilon)}{T}...(1)$

Speed=distance/time $\Rightarrow \Large S=\frac{D}{t}$

$\therefore \Large T=\frac{D}{V+\epsilon}$ and $\Large (T-1)=\frac{D}{V+\epsilon}-1$

$\therefore$ from eqn(1) Speed=$\Large \frac{D-1}{V+\epsilon}\frac{V+\epsilon}{D}(V+\epsilon)$

$\Large (1-\frac{1}{D})(V+\epsilon)$

for large D, speed=$\Large V+\epsilon$

so with the same speed you can reach your destination one minute earlier!

The last comment is a funny one, as you don't get to the destination one "earlier", you are only one minute closer to your destination compared to the start of the problem :) — Evan, Aug 22 '13 at 14:41

Collecting different answers to a simple problem

5 Answers5