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November 1st, 2019, 07:47 PM   #1
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Math Focus: Calculus
Question 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?.
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November 1st, 2019, 08:44 PM   #2
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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 $v(0)=-1$, $t= 12 , \; -8$.
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November 1st, 2019, 09:32 PM   #3
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Math Focus: Calculus
Question

Quote:
Originally Posted by tahirimanov19 View Post
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?
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November 1st, 2019, 11:44 PM   #4
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$\displaystyle a = \frac{dv}{dt}$ 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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