A car accelerates at $2m/s^2$ and decelerates at $1m/s^2$. The car starts at position $0m$ with no velocity. Calculate the shortest possible time for it to reach a distance of $100m$ without overshooting:
(a) if the maximum speed it can attain is $10m/s$
(b) if it does not have a speed limit

I have been trying to solve this but all my attempts were futile. :(

Question

A car accelerates at $2m/s^2$ and decelerates at $1m/s^2$. The car starts at position $0m$ with no velocity. Calculate the shortest possible time for it to reach a distance of $100m$ without overshooting:
(a) if the maximum speed it can attain is $10m/s$
(b) if it does not have a speed limit

I have been trying to solve this but all my attempts were futile. :(

anonymous · Answer

Can you share how u approached??? The only thing I am getting at this moment is that make it accelerate the whole journey and least time would be consumed...

anonymous · Answer

My gut sense tells me that I should approach this problem by differentiation, but I'm not sure how to start. My current approach, which is very naive, actually cuts the velocity-time graph and then slowly analyse the numbers step by step, which is not effective at all.

anonymous · Answer

Was it part (a) or (b) is 10 seconds for?

anonymous · Answer

for b part???

anonymous · Answer

I don't have the official answers for this question. But my own calculation yields 17.5s for part(a) and 5s for part(b). However, I cannot be sure of my own answer and as mentioned earlier, I'm using a very naive method. I believe this question could be approached with calculus.

anonymous · Answer

No problem, take your time to think. Thanks. :)

anonymous · Answer

For part A, the journey is 100m and goes as follows:
accelerate to 10m/s - constant velocity - decelerate to 0m/s

the first thing we need to determine is in what time it attains the speed of 10m/s:
$v=at$
$10ms^{-1}=2ms^{-2}t$
$t = 5s$

Now average speed over this 5 second interval is $\Large\frac{10m/s-0m/s}{2} = 5$m/s, so the total distance over these 5 seconds travelled is 25m

Second thing we need to consider is that it needs to stop at 100m, so it doesn't overshoot, so we need to find out how quick it stops, it decelerates at 1m/s^2, and thus stops after 10 seconds, average speed over this 10 second interval is again 5m/s, for a distance of 50m travelled. 

So with acceleration and deceleration combined you travel 75m, the remaining 25 take 2.5 sec at 10m/s, for a total travel time of 5 (acceleration phase for 25m) + 2.5 (25m at 10m/s) + 10(deceleration) = 17.5 seconds 

Part B is a bit simpler, since it doesn't have a speed limit:
Since decelerating takes twice as long as accelerating, we can cut the journey in 3 parts, the first part of it accelerating, and the second+third part decelerating:

for the first part:
$a = 2m/s^2$
$s = \frac{1}{2}at^2$
$s = t^2$ 
s = 100/3, t = $\Large \frac{10}{\sqrt{3}}$

Decelerating takes twice as long, so total time is 3t = $\Large \frac{30}{\sqrt{3}}=10\sqrt3s$

anonymous · Answer

Thanks for the help!! :D

I used the same method as you did for the first part and I got the same answer as you. For the second part, when I redo my steps again, I get my answer as $\sqrt{\frac{100}{6}} + \sqrt{\frac{200}{3}} = 12.247$. 

You stated that $s = \frac{1}{2}at^2$, but since the velocity is not known and the time isn't known too, isn't $\frac{100}{3}$ just the distance but not the time?

anonymous · Answer

Also, other than this "partitioning" method, is it possible to approach this problem with differentiation or a more systematic calculus way?

agent0smith · Answer

Maybe integrate the two velocity functions (since integral of v is distance) from 0 to a (for acceleration) and from a to t (for deceleration) and add them together, set it equal to 100, solve for a... but i just realized this may leave you with two unknowns.

agent0smith · Answer

Well I guess you can get a from knowing that the acceleration is twice the deceleration.

anonymous · Answer

For the second part, what would be the correct answer? I'm still quite confuse about the correct answer for second part.

anonymous · Answer

Well the car needs to come to a stop, so lets say it accelerates for t = x seconds:
the speed $v$ is then $\Large 2x\frac{m}{s}$, now for decelerating, we have $v = 2x - 1t$ ($v = v_0 + at$, where v0 is the speed after accelerating)
It must not overspeed, so at the end v = 0, this gives $t = 2x$ seconds,
The total time of the journey t is therefore 3x => $\Large x = \frac{t}{3}$
Now for the distance traveled, $s(t) = \frac{1}{2}at^2$ (integrate a(t) = a twice to get it.)

It traveled x seconds with an acceleration of 2m/s^2, and 2x seconds with a deceleration of 1m/s^2:

$s_{\mbox{accelerating}}(t) = \frac{1}{2}*2t^2=t^2$
$s_{\mbox{decelerating}}(t) = \frac{1}{2}*1t^2=\frac{1}{2}t^2$

The total distance traveled is:
$\Large s_{\text{acc}}(x) + s_{\text{dec}}(2x) = 100m$
->
$$\left(\frac{t}{3}\right)^2+\frac{1}{2}\left(\frac{2t}{3}\right)^2=100$$
$$\frac{t^2}{9}+\frac{1}{2}\frac{4t^2}{9}=100$$
$$\frac{t^2+2t^2}{9}=100$$
$$t^2 = 300$$
$$t = \sqrt{300} =10\sqrt{3}$$

I hope this is more clear :)

anonymous · Answer

To actually be more formal with the formulas, 
$$s_{\mbox{dec}}(t) = 2xt-\frac{1}{2}t^2$$,where 2x was the initial velocity, but I just treated it as starting at 0m/s, and accelerating at 1m/s^2 instead of starting at 2x m/s and decelerating at 1m/s^2, the result is the same tho:

$$s_{acc}(x) + s_{dec}(2x) = 100$$
$$x^2 + 4x^2 -\frac{4x^2}{2} = 100$$
$$3x^2 = 100$$
$$x = \sqrt{\frac{100}{3}}$$

x was t/3 where t was the total time the journey took thus:
$$t_{\mbox{journey}}=3\sqrt{\frac{100}{3}}=\sqrt{\frac{900}{3}}=10\sqrt3\mbox{ seconds}$$

agent0smith · Answer

^That's something like what I had in mind - find the distance travelled over both sections, added and equal to 100, knowing that the will acceleration will be 2m/s for the first third and 1m/s for the last two thirds. Nice work.