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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 $24\,m$.

Sketch of the problem

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?.

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    $\begingroup$ $v^2 - u^2 = 2ax$ gives $a = \frac{v^2-u^2}{2x} = \frac{(1+x)-1}{2x} = \frac{1}{2} $ $\endgroup$
    – AgentS
    Nov 2 '19 at 3:20
  • $\begingroup$ Looks your mistake was that missing $2$ in the denominator for acceleration $\endgroup$
    – AgentS
    Nov 2 '19 at 3:25
  • $\begingroup$ @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. $\endgroup$ Nov 2 '19 at 3:34
  • $\begingroup$ @ganeshie8 Your equation doesn't yield me some result. Perhaps can you offer an answer?. I'm stuck. $\endgroup$ Nov 2 '19 at 3:44
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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: $$a = \dfrac{1}{2}\dfrac{d}{dx}(v^2)$$

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

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  • $\begingroup$ 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. $\endgroup$ Nov 2 '19 at 4:52
  • $\begingroup$ As I mentioned above could this answer had been obtained using any calculus?. $\endgroup$ Nov 2 '19 at 4:53
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    $\begingroup$ 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. $\endgroup$
    – AgentS
    Nov 2 '19 at 4:55

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