Quadratics in Kinematics and the meaning of their solutions I have a kinematics problem, which I can happily do using the various non-quadratic kinematics equations. But when I do it with a quadratic equation, $s = vt - \frac{1}{2}at^2$, I get two possible solutions for $t$. The issue is they are both positive and there is only one physically possible solution.
The problem is this. A particle is travelling from A to B along a straight horizontal road with constant acceleration $0.34 \text{ms}^{-2}$. At B its velocity is $20\text{ms}^{-1}$ and the distance between A and B is $400\text{m}$. Find the time taken for the car to travel from A to B. The original question gives an intermediate velocity, so this can be used to avoid the quadratic equation. However, I am very bugged by why it is the case that I get the right answer 25.5 and a wrong answer, 92.1, which can't be discerned as incorrect, from the quadratic version.
I get that if it was a negative acceleration, a deceleration, that there could be a point when the particle would return and the two physically possible values would make sense. But the acceleration is in the direction of travel and the particle will move on forever. Shouldn't the incorrect solution be negative? Or is the use of the quadratic kinematic equations not advised in certain circumstances? What is the physical meaning of the larger, incorrect answer?
My workings are as follows.
$$s=vt - \frac{1}{2}at^2$$
$$400 = 20t - \frac{1}{2}(0.34)t^2$$
$$0.17t^2 - 20t +400 = 0$$
$$t = \frac{-(-20)\pm \sqrt{20^2 - 4(0.17)(400)}}{0.34}$$
$$t = 25.5  \quad \text{and} \quad t = 92.1$$
 A: The equation setup is wrong. From what I understand, the formula $s=vt-\frac{1}{2}at^2$ is used in projectile motion. This is used in the $y-$direction. If you are throwing an object, and your object is going against gravity, hence the negative acceleration (due to gravity), ie. $-a$. You can take a look on "projectile motion" on wiki for your own understanding.
You should consider the fundamental equations of motion.
$$
\begin{align}
v & = v_0 + at\\
x & = x_0+v_0t+ \frac{1}{2}at^2
\end{align}
$$
for $v_0$ and $x_0$ are your initial velocity and displacement respectively.
Now we have your above info, to find the time taken to travel $x=400m$, we do the basic substitution:
$$
\begin{align}
v & = v_0 + at \\
v_0 & = v - at \\
\end{align}
$$
We do the following algebraic substitution:
$$
\begin{align}
x & = x_0 + v_0t + \frac{1}{2}at^2 \\
& = x_0+(v-at)t+ \frac{1}{2}at^2 \\
\end{align}
$$
We substitute the values as per above:
$$
\begin{align}
400 & = 0 + (20-0.34t)t + \frac{1}{2}(0.34)t^2 \\
& = 20t - 0.17t^2 \\
\end{align}
$$
Coincidentally, we do arrive to the equation as your setup above.
Now we graph $x(t) = 20t - 0.17t^2$ for $x \in [0,400m]$ on Desmos.

Hence, when you solve your quadratic equation, the only valid solution is $t = 25.55s$.
Hope it clears your doubts.
A: Hint: Your impression that "there is only one physically possible solution" is wrong. Try working out the initial velocity $u$ for both values of $t$ and compare their signs.
A: The issue taken with the original solution seems to be confusion over "labeling" and  interpretation.  The problem as stated appears to be that the object starts at point $ \ A \ $ [position $ \ s_A \ = \ 0 \ ] \ $ with some initial velocity $ \ v_0 \ $ and travels along a straight path with a constant acceleration $ \ a \ = \ 0.34 \  \frac{\text{m}}{\text{sec}^2} \ \ , $ reaching the point $ \ B \ $ [position $ \ s_B \ = \ 400 \ ] \ $ with velocity $ \ v_B \ = \ 20 \ \frac{\text{m}}{\text{sec}} \ \ . $  The fact that the acceleration has no indicated sign is generally taken to mean that it is the same direction as distance is being measured, so the problem is stating that the speed of the object is increasing as it travels from $ \ A \ $ to $ \ B \ \ . $  The kinematic equation for distance traveled, explicitly showing time, would be
$ \ s_B - s_A \ = \ v_0·t \ + \ \frac12a·t^2 \ \ . \ $  There is of course a (temporary) difficulty in applying this, in that the initial velocity at $ \ A \ $ is not given.  We can eliminate the need to know it by using the constant-acceleration velocity equation $ \ v_B - v_0 \ = \ a·t \ $ and inserting this into the distance equation, thus
$$ 20 \ - \ v_0 \ \ = \ \ 0.34·T \ \ \Rightarrow \ \ v_0 \ \ = \ \ 20 \ - \  0.34·T $$
$$ \Rightarrow \ \ 400 \ - \ 0 \ \ = \ \ (20 \ - \  0.34·T)·T \ + \ \frac12·0.34·T^2 \ \ . \ $$
$$ \Rightarrow \ \ 400 \ \ = \ \ 20·T \ - \ \frac12·0.34·T^2 \ \ \Rightarrow \ \ 0.17·T^2 \ - \ 20·T \ + \ 400 \ \ = \ \ 0 \ \ . $$
This is the calculation Eddy Y makes, but the graph cropping is misleading, because there are two solutions for $ \ T \ $ in this kinematic equation, and they are
$$ \ T \ = \ \frac{20 \ \pm \ \sqrt{400 \ - \ 4·0.17·400}}{2·0.17} \ \ \approx \ \ 58.82 \ \pm \ 33.28 \ \ \approx \ \ 25.54 \ \ , \ \ 92.10 \ \ .   $$
[The curve for distance traveled being "concave downward" should have been concerning.  A more complete graph is shown here. This should be identified as a graph of "times at which a distance $ \ s \ $ from $ \ A \ $ is reached"; it is not depicting the trajectory followed by the object. ]

The first solution is (probably) the one that the problem-poser was looking for, which is the time it takes the object to leave $ \ A \ $ with velocity $ \ v_0 \ = \ 20 \ - \ 0.34·25.54 \ \approx \ 11.32 \ \frac{\text{m}}{\text{sec}} \ $ and increase speed at $ \ +0.34 \ \frac{\text{m}}{\text{sec}^2} \ $ toward $ \ B \ $ to arrive at $ \ B \ $ still traveling in the same initial direction $ \ 400 \ $ meters down the "road" with a speed of
$ \ 20 \ \frac{\text{m}}{\text{sec}} \ \ . $
So what about the other solution for $  \ T \ $ ?  If we return to the velocity equation, we obtain $ \ v_0' \ = \ 20 \ - \ 0.34·92.1 \ \approx \ -11.31 \ \frac{\text{m}}{\text{sec}} \ \ . \ $  So this represents the situation in which the object leaves $ \ A \ $ with speed $ \ \approx \ 11.32 \ \frac{\text{m}}{\text{sec}} \ $ headed away from $ \ B \ $ but with its acceleration of $ \ +0.34 \ \frac{\text{m}}{\text{sec}^2} \ $ toward $ \ B \  \ ; \ $ the object comes to rest momentarily after $ \ \approx \ 33.2 \ $ seconds at $ \ \approx \ 188 \ $ meters "behind" $ \ A \ \ , \ $ then increases speed toward $ \ B \ \ , \ $ passing $ \ A \ \ \approx \ 66.5 \ $ seconds after it started, and  reaching $ \ B \ $ traveling in the direction opposite to its initial velocity and now $ \ 400 \ $ meters down the "road" from $ \ A \ $ with a speed of $ \ 20 \ \frac{\text{m}}{\text{sec}} \ \ . $  The graph of these two trajectories is shown below.  (Note that the slopes of the curves at $ \ s \ = \ 400 \ $ are the same,  and the slope of the shorter path at $ \ s \ = \ 0 \ $ and of the "return" to $ \ s \ = \ 0 \ $ for the longer path are the same.)

So what's "wrong" with LostInHilbertSpace's solution?  Actually, nothing:  the difference is in the choice of "labeling".  In writing $ \ s \ = \ vt \ - \ \frac12 at^2 \ \ , \ $ and inserting $ \ s \ = \ 400 \ $ meters, $ \ v \ = \ 20 \ $ and using a "negative" acceleration term, LIHS is in a sense making a "time-reversed" calculation.  The corresponding velocity equation is $ \ v_0 - v_B \ = \ -at \ \rightarrow \ v_0 \ = \ 20 \ - \ 0.34·T \ \ , \ $ which is just the same at we used above.  It may be thought of as $ \ v_0 - v_B \ = \ a·(-t) \ \ , \ $ so that we are now counting seconds before the object arrives at $ \ B \ $ with speed $ \ 20 \ \frac{\text{m}}{\text{sec}} \ \   $  and velocity pointed from $ \ A \ $ towards $ \ B \ \ . \ $
In LIHS' version of the calculation then, the question would be interpreted as: "If an object with an acceleration of $ \ 0.34   \ \frac{\text{m}}{\text{sec}^2} \ $ directed from $ \ A \ $ to $ \ B \ $ is traveling at $ \ 20 \ \frac{\text{m}}{\text{sec} } \ $ in the direction from $ \ A \ $ to $ \ B \ $ when it reaches $ \ B \ \ , \ $ and $ \ A \ $ is $ \ 400 \ $ meters away, how long ago was the object at $ \ A \ $ ? " The rest of their computation proceeds as above, and we understand the results as
• the object departed $ \ A \ $ about $ \ 25.5 \ $ seconds earlier heading toward $ \ B \ $ with a speed of $ \ 11.3 \  \frac{\text{m}}{\text{sec} } \ \ ; \ \ $ or
• the object departed $ \ A \ $ about $ \ 92.1 \ $ seconds earlier heading away from $ \ B \ $ with a speed of $ \ 11.3 \  \frac{\text{m}}{\text{sec} } \ \ , \ $ came to rest, reversed direction, and passed $ \ A \ $ again on its way to $ \ B \ \ . \ $
(Again, the graph here is of "time elapsed" and is not a graph of trajectories.)

The trajectories would look similar to those shown on the previous graphs, but now "time-reversed", with $  \ t \ = \ 0 \ $ being the moment the object reaches $ \ B \ \ . \ $   (Here, it becomes clearer that, in a sense, both trajectories are the "same trajectory".)

Both approaches in solving the problem are acceptable, but they indicate the degree of caution that must be exercised in applying mathematical functions (and interpreting results of calculations) to assess physical situations.  (This doesn't get any better in more advanced physics.)
