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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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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 $\displaystyle a = \frac12$. Tags distance, find, object, reaches, speed, squared, time Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sanjanagaur New Users 11 November 19th, 2016 06:24 AM sanjanagaur Elementary Math 9 September 11th, 2016 12:27 PM sanjanagaur New Users 0 September 9th, 2016 01:31 PM shanti Physics 2 September 21st, 2014 09:13 PM MathsIlove Elementary Math 8 September 4th, 2014 08:28 PM

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