How to find the time after an object reaches a distance from a squared speed against distance?

The problem is as follows:

The figure from below shows the squared speed against distance attained of a car. It is known that for $$t=0$$ the car is at $$x=0$$. Find the time which will take the car to reach $$24\,m$$.

The given alternatives on my book are:

$$\begin{array}{ll} 1.&8.0\,s\\ 2.&9.0\,s\\ 3.&7.0\,s\\ 4.&6.0\,s\\ 5.&10.0\,s\\ \end{array}$$

What I attempted to do to solve this problem was to find the acceleration of the car given that from the graph it can be inferred that:

$$\tan 45^{\circ}=\frac{v^{2}\left(\frac{m^{2}}{s^{2}}\right)}{m}=1\,\frac{m}{s^{2}}$$

Using this information I went to the position equation as follows:

$$x(t)=x_{o}+v_{o}t+\frac{1}{2}at^2$$

Since it is mentioned that $$x=0$$ when $$t=0$$ this would make the equation of position into:

$$0=x(0)=x_{o}+v_{o}(0)+\frac{1}{2}a(0)^2$$

Therefore,

$$x_{o}=0$$

$$x(t)=v_{o}t+\frac{1}{2}at^2$$

From the graph I can spot that:

$$v_{o}^2=1$$

$$v_{o}=1$$

Since $$a=1$$

$$x(t)=t+\frac{1}{2}t^2$$

Then:

$$t+\frac{1}{2}t^2=24$$

$$t^2+2t-48=0$$

$$t=\frac{-2\pm \sqrt{2+192}}{2}=\frac{-2\pm \sqrt{194}}{2}=\frac{-2\pm 14}{2}$$

$$t=6,-8$$

Therefore the time would be $$6$$ but apparently the answer listed on my book is $$8$$. Could it be that I missunderstood something or what happened? Is the answer given wrong?. Can somebody help me here?.

• $v^2 - u^2 = 2ax$ gives $a = \frac{v^2-u^2}{2x} = \frac{(1+x)-1}{2x} = \frac{1}{2}$ Nov 2 '19 at 3:20
• Looks your mistake was that missing $2$ in the denominator for acceleration Nov 2 '19 at 3:25
• @ganeshie8 Where?. I can't find where is the mistake. I don't understand very well what you did. Why $v^2=1+x$? Perhaps could you develop an answer?. I look in my steps as I did used the position equation and it seems right. Nov 2 '19 at 3:34
• @ganeshie8 Your equation doesn't yield me some result. Perhaps can you offer an answer?. I'm stuck. Nov 2 '19 at 3:44

Given graph has the equation: $$v^2 = 1+x$$

Implicitly differentiate both sides with respect to $$x$$ :

$$2v\dfrac{dv}{dx} =1$$

Multiply left side by $$1=\color{blue}{\frac{dt}{dt}}$$: $$2v\dfrac{dv}{\color{blue}{dt}}\dfrac{\color{blue}{dt}}{dx}=1$$

Since $$\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{v}$$: $$2v\dfrac{dv}{dt}\dfrac{1}{v}=1 \implies \dfrac{dv}{dt}=\dfrac{1}{2}$$

In general, acceleration from $$v^2,x$$ graph is obtained by the formula: $$a = \dfrac{1}{2}\dfrac{d}{dx}(v^2)$$

• You could also implicitly differentiate with respect to $t$: $2v \frac{dv}{dt} = \frac{dx}{dt} = v$. :-)
– user657854
Nov 2 '19 at 8:01
• Wow! totally missed haha. That gives the acceleration in just one step. Thanks @EhWha :) Nov 2 '19 at 8:18

You're wrongly assuming $$\color{blue}{a=1}$$.

From the kinematics equation $$v^2 = 2\color{blue}{a}x+u^2$$, with constant acceleration,
when you graph $$v^2$$ against $$x$$, you get a linear equation of form $$y=2\color{blue}{a}x + y_0$$.
Here the slope represents $$2\color{blue}{a}$$.

$$2\color{blue}{a} = \tan(45) \implies \color{blue}{a = \frac{1}{2}}$$ Then the position function would be $$x(t)=t+\frac{1}{2}(\color{blue}{\frac{1}{2}})t^2 = t + \frac{1}{4}t^2$$

Setting that equal to $$24$$ and solving gives $$t=8$$

• I see, so it was an error from my end. So I had to consider the equation as it was given from the graph. But I wonder could this had been solved using calculus?. I thought I could do $v'=\frac{1}{2}\left(x+1\right)^{-\frac{1}{2}}$ but I was stuck on what to do with that. But it seems that it was not needed. The reason why I attempted to do that was to relate the derivative of the speed is the acceleration and from that I could find the acceleration which you referred to as $\frac{1}{2}$ but it turns out that it was not necessary as could had been found from the linear equation. Nov 2 '19 at 4:52
• As I mentioned above could this answer had been obtained using any calculus?. Nov 2 '19 at 4:53
• Absolutely, but be careful $\frac{dv}{dx}$ is not same as the acceleration $\frac{dv}{dt}$. I'll post a new answer with the calc version soon. Nov 2 '19 at 4:55