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November 1st, 2019, 02:24 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  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 $\textrm{xaxis}$ 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?. 
November 1st, 2019, 02:35 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 
Can you explain the acceleration graph. What does the dashed line from t=05 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 
November 1st, 2019, 02:51 PM  #3 
Senior Member Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 
Is it a step function? $\displaystyle a(t) = \begin{cases}0, & t<5~s \\ 2 m/s^2, & t \geq 5~s \end{cases}$ 
November 1st, 2019, 03:14 PM  #4  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
How would this affect the answer to this question?.  
November 1st, 2019, 03:17 PM  #5  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
The line seems to extend from $t\geq $5.  
November 1st, 2019, 03:25 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476 
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 $v(t) = \left(\begin{cases} 4~ &t < 5\\ 4+(t5) &5 \leq t \end{cases}\right) \dfrac m s$ $s(t) = \left(\begin{cases} 4t &0\leq t < 5\\ (t5)^2 + 4(t5) + 20 &5 \leq t \end{cases}\right)~m$ you should be able to answer from this. 
November 1st, 2019, 05:14 PM  #7  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Will this update change the conditions? Quote:
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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acceleration, displacement, find, graph, object, speed 
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