Collecting different answers to a simple problem 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.

 A: 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.
A: 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

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, $t = V\Delta/\epsilon$.
A: 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, $t\epsilon=V\delta$.
A: 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). 
A: 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!
