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October 21st, 2019, 06:33 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  How do I find the displacement of a particle at a given instant?
The problem is as follows: A particle is moving along a straight line at a constant acceleration of $3 \frac{m}{s^{2}}$. If it is known that an instant of $t=4\,s$ its displacement is $100\,m$. If is also known when $t=6\,s$ its speed is $15\,\frac{m}{s}$. What will be its displacement on that instant?.$\begin{array}{ll} 1.&\textrm{128 m}\\ 2.&\textrm{130 m}\\ 3.&\textrm{144 m}\\ 4.&\textrm{124 m}\\ 5.&\textrm{152 m} \end{array}$ For this particular problem I attempted to use the motion equation as follows: $x(t)=x_{o}+v_{ox}t+\frac{1}{2}at^2$ Then I thought that the reference point would be $x_{o}=0$ and from the given conditions ($t=4$, $x=100$, $a=3$) it could be found the initial speed as follows: $x(t)=0+v_{ox}t+\frac{1}{2}at^2$ $x(t)=x_{o}+v_{ox}t+\frac{1}{2}at^2$ $100=x(4)=4v_{ox}+ \frac{3\times 4 \times 4}{2}$ $100=4v_{ox}+ \frac{3\times 4 \times 4}{2}$ $25=v_{ox}+ 6$ $v_{ox}=19$ Using this known speed and the other known speed as 15 \frac{m}{s} and the elapsed time between the two I could calculate the "acceleration" for that given period. I assumed that it was during this part the object is slowing down or deccelerate hence $a$ would be negative. (Note: For the sake of brevity and understanding I´m omitting the units but all are accordingly and consisten) $v_{f}=v_{o}+at$ $v_{f}=19+3(4)=31$ $15=31+a(2)$ $a=\frac{1531}{2}=8$ $v_{f}^{2}=v_{o}^{2}+2 a \Delta x$ $ \left(15\right)^{2} = \left(31\right )^{2}+2 \left(8\right) \Delta x $ $2\left(8\right) \Delta x = 225961 = 736 $ $ \Delta x = 46 $ Therefore wouldn't its displacement to that instant be: $100+46= 146 m$. However this answer does not appear in the alternatives. What could I be possibly doing wrong?. Can somebody guide me?. I tried to look this question in different ways and still I can't get to an answer. By the way I did transcribed from the original source and made sure that all details are right. But I find a bit misleading wouldn't the displacement at a specific instant be zero?. Probably the intended meaning was the displacement until that instant but I am still confused. Can somebody help me with this?. 
October 21st, 2019, 08:15 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677  Quote:
$\displaystyle \frac{dv}{dt} = 3$ $\displaystyle v(t) = 3t + C$ $v(6) = 15 \implies C = 3$ $\displaystyle \frac{dx}{dt} = 3t3$ $x(t) = \dfrac{3t^2}{2}  3t + C$ $x(4) = 100 \implies C = 88$ $x(t) = \dfrac{3t^2}{2}  3t + 88$ $x(6) = 124$  
October 21st, 2019, 08:57 PM  #3  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Gee How could I overlooked integrals Quote:
 
October 22nd, 2019, 05:36 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677  Quote:
$v(6)=v(4)+3(2) \implies v(4)= 9$ $x(6) = x(4) + v(4) \cdot 2 + \dfrac{1}{2}(3) \cdot 2^2$ $x(6) = 100 + 18 + 6$  

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displacement, find, instant, particle 
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