How can I find time in physics with a motion problem? My problem is:
A car slows down at $-5.00~\text{m}/\text{s}^2$ until it comes to a stop after travelling $15.0~\text{m}$. How much time did it take to stop? (unit $= s$)
So this is how my school taught me to solve this:
\begin{align*}
t & = ?\\
a & = -5.00~\text{m}/~\text{s}^2\\
v_f & = 0\\
\delta x & = 15.0
\end{align*}
$$\delta x = v_f t - \frac{1}{2}at^2$$ 
(missing $v_i$)
I have tried to find time using distance over speed, but that answer is incorrect. Is there a way to solve this problem using the formula my school gave me (rearranging the formula using algebra), or is there another way? Thanks!
 A: Your information is the constant acceleration
$$
\ddot{x}(t) = - 5 \frac{\text{m}}{\text{s}^2}
$$
Integration regarding $t$ and assuming $t_0 = 0\, \text{s}$ gives the velocity
$$
\dot{x}(t)  = -5 \frac{\text{m}}{\text{s}^2} t +  \dot{x}_0
$$
where $\dot{x}_0 = \dot{x}(t_0)$.
Integrating again and assuming $x(t_0) = 0\,\text{m}$ gives the position
$$
x(t) = 
-\frac{5}{2} \frac{\text{m}}{\text{s}^2} t^2 + 
\dot{x}_0 t
$$
Coming to a stop means
$$
\dot{x}(t) = 0 \frac{\text{m}}{\text{s}} 
$$
or
$$
t = \frac{\dot{x}_0}{5 \text{m}/\text{s}^2}
$$
So we have
$$
\begin{align}
15\,\text{m} = x(t) 
&= 
-\frac{5}{2} \frac{\text{m}}{\text{s}^2} \left( \frac{\dot{x}_0}{5 \text{m}/\text{s}^2} \right)^2 + \dot{x}_0 \frac{\dot{x}_0}{5 \text{m}/\text{s}^2} \\
&= \left( -\frac{1}{10\,\text{m}/\text{s}^2} + \frac{1}{5\,\text{m}/\text{s}^2}\right) \dot{x}_0^2 \\
&= \frac{1}{10\,\text{m}/\text{s}^2}  \dot{x}_0^2 \iff \\
\dot{x_0} &= \sqrt{150} \frac{\text{m}}{\text{s}}
\approx 12.25 \frac{\text{m}}{\text{s}}
\end{align}
$$
and thus
$$
t = \frac{\sqrt{150}}{5}\,\text{s} = \sqrt{6} \,\text{s} \approx 2.5 \,\text{s}
$$
Here is a graph:

A: The equation 
$$\delta x = v_f t - \frac{1}{2}at^2$$
is a quadratic equation in $t$, which can be expressed in the form
$$\frac{1}{2}at^2 - v_f t + \delta x = 0$$
so it could be solved by using the Quadratic Formula.  
However, in this case, we know that $v_f = 0$.  Hence,
$$\delta x = -\frac{1}{2}at^2$$
Solving for $t$ yields
\begin{align*}
-2\delta x & = at^2\\
-\frac{2\delta x}{a} & = t^2\\
\sqrt{-\frac{2\delta x}{a}} & = |t|\\
\sqrt{-\frac{2\delta x}{a}} & = t && \text{since $t \geq 0$}
\end{align*}
You can now substitute $15.0~\text{m}$ for $\delta x$ and $-5.00~\dfrac{\text{m}}{\text{s}^2}$ for $a$ to solve for $t$. 
A: Tip-
Your method went wrong because you used $t=s/v$, where $t=$time, $s=$distance,$v=$speed incorrectly.
Here, your car is accelarating, which means that there is no constant speed that you can use for the above formula (You can use the average speed though).

Solution-
You have -
Final speed, $v=0ms^{-1}$,
distance travelled, $s=15m$
and, accelaration $a=(-5)ms^{-2}$,
You have to find -
Time taken, $t=?$
The given equation, $s=vt-{1 \over 2} a t^2$, is a quadratic equation in t.
Substituting the above values in it gives-
$15=0+{5 \over 2}t^2$, (because, $v=0$ and $a=(-5)$as is given in the question)
Giving, time $t=\pm \sqrt{6}s$
Since, time cannot be negative your answer is $\sqrt{6}s$.  (Answer)
A: I don’t know if you’ve covered this yet, but a convenient way to solve problems like this one is to use conservation of energy. The kinetic energy of the the car is equal to $\frac12mv^2$. The change in its energy is equal to the work done on it. Since the acceleration $a$ is constant, then so is the force, and the work is thus equal to $mas$, where $s$ is the distance traveled. Therefore, you have $$\frac12mv_f^2-\frac12mv_i^2=mas,$$ from which $$v_f^2-v_i^2 = 2as.\tag1$$ On the other hand, because of constant acceleration, you have $$v_f-v_i = at.\tag2$$ In this problem, $v_f=0$. Squaring (2) and substituting for $v_f$ in (1) gives you $-(at)^2 = 2as$, which you can then solve for $t$ to get $$t = \sqrt{-2s/a}.$$
A: It is an easy kinematics problem from physics. Write down motion equations.
$ 0=V_0-at(1)$
$x  =V_0t-\frac{at^2}{2} (2)$ 
from (1) $V_0=at (1')$
Put (1') to (2) $x=\frac{at^2}{2}$ or $t=\sqrt{\frac{2x}{a}}=\sqrt{\frac{2*15}{5}} = \sqrt{6}$
No pictures or quadratic solutions or kinetic energy formulas are required. Just physics.
