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.
$$s=vt – \frac{1}{2}at^2$$
$$400 = 20t – \frac{1}{2}(0.34)t^2$$
$$0.17t^2 – 20t +400 = 0$$
$$t = \frac{-(-20)\pm \sqrt{20^2 – 4(0.17)(400)}}{0.34}$$
$$t = 25.5 \quad \text{and} \quad t = 92.1$$
Best Answer
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.