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 November 1st, 2019, 07:47 PM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus How to find the time after an object reaches a distance from a squared speed? 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: $\tan 45^{\circ}=\frac{\frac{m^{2}}{s^{2}}}{m}$ $\tan 45^{\circ}=1\,\frac{m}{s^{2}}$ 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?.
 November 1st, 2019, 08:44 PM #2 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle What we know: $x(0)=0$ and $v(0)= \pm 1$. For constant acceleration $a= \dfrac{v^2(t) - v^2(0)}{2(x(t)-x(0))}$ From graph $v^2=x+1$ so $a=\dfrac{(x+1)-1}{2x} = 1/2$ $x(t) = v(0)t+ \dfrac{1}{2}at^2= \pm t + \dfrac{t^2}{4}$ So, $24=\pm t + \dfrac{t^2}{4} \Rightarrow t^2 \pm 4t -96=0$ For $v(0)=1$, $t= -12 , \; 8$. For $v(0)=-1$, $t= 12 , \; -8$.
November 1st, 2019, 09:32 PM   #3
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Joined: Jun 2017
From: Lima, Peru

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Math Focus: Calculus

Quote:
 Originally Posted by tahirimanov19 What we know: $x(0)=0$ and $v(0)= \pm 1$. For constant acceleration $a= \dfrac{v^2(t) - v^2(0)}{2(x(t)-x(0))}$ From graph $v^2=x+1$ so $a=\dfrac{(x+1)-1}{2x} = 1/2$ $x(t) = v(0)t+ \dfrac{1}{2}at^2= \pm t + \dfrac{t^2}{4}$ So, $24=\pm t + \dfrac{t^2}{4} \Rightarrow t^2 \pm 4t -96=0$ For $v(0)=1$, $t= -12 , \; 8$. For $v(0)=-1$, $t= 12 , \; -8$.

From the graph, wouldn't it be that the $\textrm{acceleration = 1}$?

I came to that conclusion because:

$\tan 45^{\circ}=1$

When you use the equation for obtaining the acceleration from initial and final speeds, shouldn't it check with $1$ but why $\frac{1}{2}$?

This part is where I'm confused.

Because if $a=\frac{1}{2}$ the rest is like domino pieces and it is not hard to understand but I'm still confused on where does this "half" appears. Can you help me with a clarification for that part?

Does it exist another way to obtaining the same value for the acceleration, perhaps using derivatives?

 November 1st, 2019, 11:44 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 $\displaystyle a = \frac{dv}{dt}$ and $\displaystyle v = \frac{dx}{dt}\\$. If we assume $v^2 = x + 1$ for $x \geqslant 0\\$, differentiating w.r.t. $t$ gives $\displaystyle 2v\frac{dv}{dt} = \frac{dx}{dt}$, so $2va = v$. Except for $v = 0$, dividing by $2v$ gives $\displaystyle a = \frac12$.

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