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November 1st, 2019, 02:24 PM   #1
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Question How can I find the speed and the displacement of an object from acceleration graph?

The problem is as follows:
An object is moving along the $\textrm{x-axis}$ with an initial speed of $+4\hat{i}\,\frac{m}{s}$, its acceleration against time is given in the graph from below. Find the speed on $\frac{m}{s}$ of the object for $t=9\,s$ and its displacement (on $m$) during the first $8\,s$.


The given alternatives are:

$\begin{array}{ll}
1.&+12\hat{i}\,\frac{m}{s}\,;52\,m\\
2.&+12\hat{i}\,\frac{m}{s}\,;41\,m\\
3.&+14\hat{i}\,\frac{m}{s}\,;41\,m\\
4.&+16\hat{i}\,\frac{m}{s}\,;52\,m\\
\end{array}$

This problem looks trivial but I'm stuck with the second part which is to find the displacement as for me it doesn't look very obvious. Can somebody help me with this?

What I did for the first part was to use this formula:

$v_{f}=v_{o}+at$

Since the initial speed is given and the acceleration is $a=2$ then:

$v_{f}=4+(2)(4)=12\,\frac{m}{s}$

But the displacement is where I'm stuck at:

Wouldn't it be:

$v_{f}^2=v_{o}^2+2a\Delta x$

$12^2=4^2+2(2)\delta x$

$\delta x = \frac{144 - 16}{2}=\frac{128}{2}=64 m$

But this does not check with any of the alternatives given. What should I do?.
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November 1st, 2019, 02:35 PM   #2
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Can you explain the acceleration graph.

What does the dashed line from t=0-5 stand for?
What does the solid line for t>5 stand for?

I'm assuming from the graph that acceleration is constant at $2~\dfrac{m}{s^2}$ for $5 \leq t$
but I really don't know what the dashed line from $0\leq t < 5$ means
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November 1st, 2019, 02:51 PM   #3
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Is it a step function?
$\displaystyle a(t) = \begin{cases}0, & t<5~s \\ 2 m/s^2, & t \geq 5~s \end{cases}$
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November 1st, 2019, 03:14 PM   #4
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Quote:
Originally Posted by romsek View Post
Can you explain the acceleration graph.

What does the dashed line from t=0-5 stand for?
What does the solid line for t>5 stand for?

I'm assuming from the graph that acceleration is constant at $2~\dfrac{m}{s^2}$ for $5 \leq t$
but I really don't know what the dashed line from $0\leq t < 5$ means
I just copied it down as I see it on my pamphlet. I believe the intended meaning was that the object starts moving at $5\,s$ and was at rest for the time before.

How would this affect the answer to this question?.
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November 1st, 2019, 03:17 PM   #5
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Quote:
Originally Posted by DarnItJimImAnEngineer View Post
Is it a step function?
$\displaystyle a(t) = \begin{cases}0, & t<5~s \\ 2 m/s^2, & t \geq 5~s \end{cases}$
Judging from what it was intended in the original source. It "seems" to have the shape of one, but will this affect the answer?.

The line seems to extend from $t\geq $5.
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November 1st, 2019, 03:25 PM   #6
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Ok assuming it means an acceleration of $a=2~\dfrac{m}{s^2}$ turns on at $t=5~s$, and assuming $v(t) = 4~\dfrac m s$ for $t < 5$, and finally assuming that $s(0) = 0$

we have

$v(t) = \left(\begin{cases}
4~ &t < 5\\
4+(t-5) &5 \leq t
\end{cases}\right) \dfrac m s$

$s(t) = \left(\begin{cases}
4t &0\leq t < 5\\
(t-5)^2 + 4(t-5) + 20 &5 \leq t
\end{cases}\right)~m$

you should be able to answer from this.
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November 1st, 2019, 05:14 PM   #7
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Lightbulb Will this update change the conditions?

Quote:
Originally Posted by romsek View Post
Ok assuming it means an acceleration of $a=2~\dfrac{m}{s^2}$ turns on at $t=5~s$, and assuming $v(t) = 4~\dfrac m s$ for $t < 5$, and finally assuming that $s(0) = 0$

we have

$v(t) = \left(\begin{cases}
4~ &t < 5\\
4+(t-5) &5 \leq t
\end{cases}\right) \dfrac m s$

$s(t) = \left(\begin{cases}
4t &0\leq t < 5\\
(t-5)^2 + 4(t-5) + 20 &5 \leq t
\end{cases}\right)~m$

you should be able to answer from this.
Since apparently there was some sort of confusion. I'm adding this new graph which points a missing element from the original which was a tiny white circle over the number $5$.



I'm not very sure about this. But can we say beforehand that the object is at rest before $t=5$?. Is this a valid conclusion?. On which sort of graphic we can do such assumption?.

I was reading at your solution, but mind guiding me on how you obtained the $c=20$? As I assume the equations for the position you obtained them by integrating the speeds. Am I right?.

Last edited by Chemist116; November 1st, 2019 at 05:18 PM. Reason: Added necesary clarification
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