Calculate time to cover distance with accelaration and decelaration

For a project I am doing, I need to calculate the time a car will take to travel from point A to point B, given that:

• The distance between A and B is straight
• The car has starting velocity Vstart, and a maximum velocity Vmax
• At point B, a curve starts, where the car must have a different maximum velocity: Vcurvemax. This velocity is typically lower than Vmax
• The car will accelerate coming from A, at its accelaration rate, at most until it reaches velocity Vmax. At some point along the line from A to B, whether or not Vmax was reached, the car needs to decelarate (at its decelaration rate), in order to be at velocity Vcurvemax when reaching point B

My question is: is it possible to calculate precisely the time the car will take to cover the distance between A and B, given these circumstances? Actually, this comes down to another question: is it possible to calculate precisely the point between A and B where the car needs to start decelarating?

I could estimate it fairly accurately by doing a few calculations (in a programmatic loop) to find out just about where the car will have to start decelarating. However, I wonder if a calculation could precisely determine this.

Thanks for any ideas!

• Is there a maximum acceleration and deceleration limit? If no, theoretically, we can have infinite acceleration at a short time to increase the velocity from initial to max, in order to use least time to travel AB. And then have infinite deceleration at point B to decrease the velocity to the new max. – Szeto Feb 16 '18 at 11:29
• The acceleration and decelaration are seperately defined, but fixed, for example at 2,5 m/s2. They will, in my case, never be infinite. Depending on the distance between A and B (and the initial velocity), Vmax may or may be reached before the car has to start decelaration in order to be at Vmaxcurve at point B. – mennowo Feb 16 '18 at 12:22

The given data are the initial velocity $v_A$, the end velocity $v_B$, the maximal speed $\bar v$, the maximal acceleration $a^\uparrow>0$, the maximal deceleration $a^\downarrow>0$, and the distance $d=|AB|$. From these data the time needed to travel from $A$ to $B$ is determined, and can be simply calculated. Three different cases can arise, depending on the numerical values of the data:

(i) If $v_A$ is too large in comparison with $d$ and $a^\downarrow$ the trip cannot be done as required.

(ii) If case (i) does not hold we may accelerate with rate $a^\uparrow$ to a certain maximal speed $v_0<\bar v$, then decelerate to $v_B$ with rate $a^\downarrow$.

(iii) If $d$ is even larger in comparison we may accelerate with rate $a^\uparrow$ until the maximal speed $\bar v$ is reached, then cruise at that speed until a well defined time point, where we start decelerating at rate $a^\downarrow$ and arrive at $B$ with the required speed $v_B$.

Concerning case (ii):

At the beginning one has $$v(t)=v_A+a^\uparrow t,\quad s(t)=v_A t+a^\uparrow{t^2\over2}\ .$$ The state $v(t)=\bar v$ is reached at time $t_1={\bar v-v_A\over a^\uparrow}$, and the necessary length of road to reach this state is $$s_{\rm acc}=v_A t_1+a^\uparrow{t_1^2\over2}\ .$$ Similarly a full decelerating process takes time $t_2={\bar v-v_B\over a^\downarrow}$ and requires $$s_{\rm dec}=v_B t_2+a^\downarrow{t_2^2\over2}$$ of road length. If $s_{\rm acc}+s_{\rm dec}>d$ then we cannot accelerate to the maximal possible speed, and we are in case (ii). In this case one has to solve the system of equations $$v_A t'+a^\uparrow{t'^2\over2}+v_B t''+a^\downarrow{t''^2\over2}=d,\qquad v_A+a^\uparrow t'=v_B+a^\downarrow t''\ .$$ The total time then is $t'+t''$.

• Hi @Christian, thank you for your concise and very clear summary of the problem. Situation (i) and (iii) i can solve. Also I can determine when situation (ii) applies. However, I find it hard to wrap my head around what the 'certain maximal speed' would be, and when/where it would occur, which would allow calculating the time needed. – mennowo Feb 19 '18 at 14:38

HINTS

Use the following for acceleration $a$ constant

• $v=v_0+at$
• $x=x_0+v_0t+\frac12at^2$

No it is not possible.

If you consider a velocity time graph, with an initial velocity, maximum velocity and final velocity specified, but not the times at which the maximum or final velocity occurs, or the distance at which the maximum occurs, there are infinitely many curves or sets of straight lines which will have the same area under the graph which represents the fixed distance $AB$.

• The point at which the final velocity (Vcurvemax) occurs is defined as point B. If we fix acceleration and deceleration at some value (independent of on another), will there not be only a single definitive solution where the car first accelerates, then (maybe, if the distance and Vstart allow for this) stays at Vmax a little, then decelerates to reach Vcurvemax at point B? – mennowo Feb 16 '18 at 12:26
• There are still infinitely many constructions from these data – David Quinn Feb 16 '18 at 13:02