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Trials are being undertaken on a horizontal road to test the performance of an electrically powered car. The car has a top speed of $V$. During a test run, the car moves from rest with uniform acceleration $a$ and brought to rest with uniform retardation $r$.

If the car is to achieve its maximum speed $V$ the length of a test run must be at least $$\frac{V^2(a+r)}{2ar}$$

What is the least time for a test run of length $$\frac{2V^2(a+r)}{9ar}$$ ?

It should be $\frac{2V(a+r)}{3ar}$

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    Hint: assume that the car accelerates for a distance $s$ and then immediately switches to deceleration. Find the value of $s$ for which the speeds at the start and end are zero. I don’t expect that they want you to also prove that accelerating for as long as possible will yield the shortest transit time. – 2017-01-13
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    Thanks, I'll try that. I have corrected the equation above to $\frac{2V(a+r)}{3ar}$ – 2017-01-13

2 Answers 2

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Minimizing transit time is equivalent to maximizing average speed over the run. This in turn means that the car must accelerate for as long as possible—coasting or decelerating doesn’t increase the average speed. The car will then decelerate to a stop over the remaining distance.

Direct solution: The length of the track is given as $L=\frac29{a+r\over ar}V^2$, with $V$ equal to the car’s top speed. We proceed by finding the distance $s$ at which the car must begin decelerating. By conservation of energy, we have $as=r(L-s)$, so $s={r\over a+r}L=\frac29{V^2\over a}$, with the remainder equal to $(L-s)={a\over a+r}L=\frac29{V^2\over r}$. This makes sense: kinetic energy is added or shed linearly with distance, so we expect the total distance to be divided up proportionally to the acceleration and deceleration. Under constant acceleration with initial velocity $0$, $s=\frac12at^2$, so the time to travel some distance $s$ is $\sqrt{2s/a}$ and thus the total time for the run is $$\sqrt{2s/a}+\sqrt{2(L-s)/r} = \sqrt{\frac49{V^2\over a^2}}+\sqrt{\frac49{V^2\over r^2}} = \frac23\left(\frac1a+\frac1r\right)V = \frac23{a+r\over ar}V.$$

Shortcut solution: Invert the expression for track length to find the maximum velocity for a given track length $l$, i.e., $v^2=2{ar\over a+r}l$. Plugging in the given track length, $$v^2=2{ar\over a+r}\cdot\frac29{a+r\over ar}V^2=\frac49V^2$$ therefore the maximum speed for this run is $\frac23V$. Under uniform acceleration the average speed over an interval is the average of the starting and ending speeds (giving the same average speed over both segments), and the time taken is simply distance over speed, which gives $$\frac29{a+r\over ar}V^2\Big/\frac V3 = \frac23{a+r\over ar}V$$ for the run time.

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    A good response for the direct solution, thanks. Using $s = ut + \frac{1}{2} at^2$ we have $u=0$, for the acceleration stage, but for the deceleration stage we can still say that $s=\frac{1}{2} rt^2$, where the $ut$ term is also $0$, (even though the car is travelling at a speed V at the start of the deceleration stage), as we have constant deceleration, from which you got $\sqrt\frac{2(L-s)}{r}$ for the deceleration time. – 2017-01-15
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    Looking at the whole test in its entirety, from the start to the finish, the initial velocity $u=0$, when we start the acceleration stage. So this applies all the way through our calculations to the end of the deceleration stage. – 2017-01-15
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    The methods do not seem to work for a test length of $$\frac{2V^2(a+r)}{3ar}$$ which should turn out to be $$\frac{7V(a+r)}{6ar}$$ but using either method results in $$\frac{2\sqrt3V(a+r)}{3ar}$$ The test length of $$\frac{2V^2(a+r)}{3ar}$$ is 3 times the length of $$\frac{2V^2(a+r)}{9ar}$$. – 2017-01-15
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    @J132 Of course not. This new track is longer than the minimum length required to reach its top speed, so the assumption that the car is either accelerating or deceleration the entire time breaks down. You also have to take into account that the car now maintains a constant speed over some part of the track: it accelerates over a distance of $V^2/2a$, decelerates over a distance of $V^2/2r$ and travels at a constant speed of $V$ over the remaining distance. So, the time taken is $V(a+r)/ar+(L-L_{\text{min}})/V)$. – 2017-01-15
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    @J132 BTW, with the way that the parameters of the problem are given, the strictly correct model for the deceleration phase is $s=s_{\text{cutover}}+v_{\text{max}}t-rt^2/2$, but we can reverse both time and distance for the simpler $rt^2/2$. – 2017-01-15
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To show that the total time, $T$, taken to test a car on a test track with a length of $\frac{2V^2(a+r)}{3ar}$ is $\frac{7V(a+r)}{6ar}$:-

In the acceleration phase, using $v^2 = u^2 + 2as$, where $u=0$ (as before), and $v=V$ (the max speed) and $s=l$ (the length of the track for the acceleration phase of the test), so we get $$V^2 = 2al$$ and so $$ l=\frac{V^2}{2a}$$

In the deceleration phase, using $v^2 = u^2 + 2as$, where $u=0$, $v=V$ and $s=L$ (the length of the track for the deceleration phase of the test), so we get $$V^2 = 2rL$$ and so $$ L=\frac{V^2}{2r}$$

Now using $s= ut +\frac{1}{2}at^2$, with $u=0$, we get $t_a=\frac{V}{a}$, for the acceleration phase and $t_r=\frac{V}{r}$, for the deceleration phase.

So the total time for these 2 phases is: $$ t_a + t_r = \frac{V}{a} + \frac{V}{r} = \frac{V(a+r)}{ar}$$

Now we also have a constant speed phase, which is after the acceleration phase and before the deceleration phase and is attained when the min distance of the track has been driven i.e. $\frac{V^2(a+r)}{2ar}$, so the constant speed phase of the test occurs at $(s - s_{min})$ (the total distance of the track minus the min distance). Using $s = ut +\frac{1}{2}at^2$, with $u=V$ and $a=0$, we have $$(s - s_{min})=Vt_c$$ so the time, $t_c$, for the constant speed phase of the test occurs at $$t_c=\frac{(s - s_{min})}{V} = \frac{2V(a+r)}{3ar} - \frac{V(a+r)}{2ar} = \frac{V(a+r)}{6ar}$$

Now, the total time $T$ is $$T = t_a + t_r + t_c = \frac{V(a+r)}{ar} + \frac{V(a+r)}{6ar} = \frac{7V(a+r)}{6ar}$$