A car moves from point $A$ to $B$ in $t_{\text{total}}$.
$$v_i=0\ m/s\\
a= 0.65\ m/s^2\ (acceleration)\\
d=-0.65\ m/s^2\ (deceleration)\\
v_{max} = 1 m/s\\
t_{\text{total}}=10.0\ s\\
s_{A\to B}(\text{Distance})=15.0\ m\\$$
Problem 1
Solve $t_{\text{acc}}$ when the car does not decelerates towards point
$B$.In other words, how long can the car accelerate until it
reaches the velocity that will get the car to point $B$ at
$t_{\text{total}}$?Behaviour of the car:
The car will accelerate to $v_{\text{max}}$ and then keep this velocity until it reaches point $B$.
I am aware of the fact the car will 'overshoot' point $B$ in this case.
Edit for problem 1 (@Botond)
$$
\text{Expressing the problem in distances:}\\
\text{We know:}\\
$$
$$
s = v_f(t_{total} – t_{acc})\\
v_f = v_i + at
$$
$$
\text{We have 2 equations}\\
s_{acc}= \frac{v_i+v_f}{2}t_{acc}\\
s_{const} = v_f(t_{total} – t_{acc})\\
$$
$$
\text{Combine the 2 equations:}\\
t_{acc} . a . (t_{total} – t_{acc}) + \frac{t_{acc}.a}{2} . t_{acc} = S_{A\to B}\\
t_{total}.t_{acc}.a.-at_{acc}^2 + \frac{1}{2}at_{acc}^2 = S_{A\to B}\\
-\frac{1}{2}at_{acc}^2+t_{total}.at_{acc}=S_{A\to B}\\
\text{Can write as:}\\
-\frac{1}{2}ax^2+Tax=S_{A\to B }\\
\text{where:}\\
x = t_{acc}\\
T = t_{total}\\
$$
$$
\text{Now with actual values:}\\
$$
$$
T = 10sec.\\
S = 15 m.\\
a = 0.65 m/s^2\\
-0.325^2+(10 * 0,65)x=S_{A\to B }\\
-0.325x^2+6,5x=S_{A\to B }\\
\text{This leaves me the quadratic problem:}\\
-0.325x^2 + 6,5x = 10\\
-0.325x^2 + 6,5x – 10 = 0\\
\text{solves}\\
x_1 = 2.6620061429465727\\
x_2 = 17.337993857053426\\
x_1 \text{seems to be the right answer, I will edit this post to prove it}
$$
Problem 2
Solve $t_\text{acc}$ and $t_{\text{dec}}$ when the car comes to a
complete stop at point $B$.Behaviour of the car:
The car will accelerate to $v_{\text{max}}$ and start deceleration at $t_{\text{2}}$ so it will come to a stop at $B$.
If the car can reach $v_{\text{max}}$ it will accelerate to $v_{\text{max}}$ and reach this velocity at $t_{\text{1}}$. Then at $t_{\text{2}}$ it needs to decelerate to $v=0$ so it will have $v=0$ at point $B$.
If it can not reach $v_{\text{max}}$ then I want to calculate $t_{\text{1}}$ where we the acceleration flips to deceleration.
Edit for problem 2 (@Botond)
Try to express distance traveled:
Values given to solve this problem:
$$
a = 0.65 m/2^2\\
d = 0.65 m/s^2\\
v_{max} = 1.0 m/s\\
t_{total} = 10.0 s\\
s_{A\to B} = 15.0 m\\
$$
The problem states insinuates we do not know if the car will reach $v_{max}$ in the given time, or if the trajectory is even possible in the given time. This will be my first check, do we reach $v_{max}$. If we do reach $v_{max}$ we have a slightly different calculation than if we do not reach $v_{max}$.
We calculate the time it would take the car to accelerate from $v_{i} = 0$ to $v_{max} = 1.0 m/s$ and decelerate back to $v_{f} = 0$ with $a = d = 0.65m/s^2$.
$$
t_{acc} = v_{max}/a \\
t_{dec} = v_{max}/d \\
t_{ad} = t_{acc} + t_{dec} \\
$$
$$
t_{acc} = 1.0 / 0.65 \\
t_{acc} = 1.5384615385 s\\
t_{dec} = 1.0 / 0.65 \\
t_{dec} = 1.5384615385 s\\
t_{ad} = 1.5384615385 s + 1.5384615385 s\\
t_{ad} = 3.076923077 s\\
$$
Now we know the time it will take a complete acceleration and deceleration. We can calculate the displacement for $t_{ad}$. In case $a == d$ we can use $t_{ad}$. In case $a != d$ we should use $t_{acc}$ and $t_{dec}$.
For $a == d$:
$$
s = 0.5at^2\\
s_{ad} = at_{ad}^2\\
s_{ad} = 0.65 * 3.076923077^2\\
s_{ad} = 6.1538461542 m
$$
For $a != d$
$$
s = 0.5at^2\\
s_{ad} = 0.5at_{acc}^2 + 0.5at_{dec}^2\\
s_{ad} = 0.325 (3.076923077^2) + 0.325 (3.076923077^2)\\
s_{ad} = 6,1538461542 m
$$
Now we know the time it will take the car to reach $v_{max}$ and back to $v_{f} = 0$. If we compare $t_{ad}$ with $t_{total}$ we say:
If $t_{total} < t_{ad}$ the car will not reach $v_{max}$
If $t_{total} >= t_{ad}$ the car will reach $v_{max}$
For both situations applies different logic.
Situation $t_{total} >= t_{ad}$:
There are three parts in this trajectory. The acceleration part, the constant velocity part, and the deceleration part. For each part we can now calculate the displacements:
$$
s_{acc} = 0.5at^2\\
s_{acc} = 0.325(t_{acc}^2)\\
s_{acc} = 0.325(t_{acc}^2)\\
s_{acc} = 0.325(2.3668639054)\\
s_{acc} = 0,7692307693\\
$$
$$
t_{acc} = \frac{v_{const}-v_i}{a}\\
t_{dec} = \frac{v_f – v_{const}}{a}\\
t_{const} = t_{total} = t_{acc} – t_{dec}\\
$$
$$
s_{acc} = \frac{v_i+v_{const}}{2}t_{acc}\\
s_{dec} = \frac{v_{const}+v_f}{2}t_{dec}\\
s_{const} = v_{const}.t_{const}\\
$$
$$
\text{Combine the 3 distance equations:}\\
v_f(t_{const}) + \frac{v_i + v_{const}}{2}.t_{acc} + \frac{v_{const} + v_f}{2}.t_{dec} = s_{A\to B}\\
\text{Since:}\\
v_f = v_i + at\\
\text{we can rewrite to:}\\
(v_i + at)(t_{const}) + \frac{v_i + v_{const}}{2}.t_{acc} + \frac{v_{const} + v_f}{2}.t_{dec} = s_{A\to B}\\
$$
Work in progress…
Best Answer
For problem 1: The car is accelerating for $t_{acc}$ time and going with the same velocity for the rest of the time: $t_{total}-t_{acc}$. How much distance will it cover during the acceleration? And what about the rest? I think you can calculate those easily. And their sum must be the $s_{A \to B}$.
For problem 2: Try to express the distance traveled, just like in the first problem.
Some useful formulas: $$s=v_it+\frac{a}{2}t^2$$ $$s=\frac{v_i+v_f}{2}t$$ $$v_f=v_i+at$$ $$v_f=\sqrt{2as+v_i^2}$$ For the first one: We are accelerating to $v_{f}$ from $v_i=0$. It will take $$t_{acc}=\frac{v_f}{a}$$ time. In the second part of te trip, we are going with constant $v_f$ velocity for $t-t_{acc}$ time. The distance traveled during this time is $$s=v_f(t_{total}-t_{acc})$$ So we have $2$ equations: $$v_f(t_{total}-t_{acc})+\frac{v_f}{2}t_{acc}=s_{A \to B}$$ $$t_{acc}=\frac{v_f}{a}$$ Which implies that $$v_f\left(t_{total}-\frac{v_f}{a}\right)+\frac{v_f}{2} \frac{v_f}{a}=s_{A \to B}$$ Or $$t_{acc}a(t_{total}-t_{acc})+\frac{t_{acc}a}{2}t_{acc}=s_{A \to B}$$