How can I find time in physics with a motion problem?

Question

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!

Answer 1

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 $t$.

Answer 2

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).

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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)

Answer 3

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:

Answer 4

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 $$t = \sqrt{-2s/a}.$$

Answer 5

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.