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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) $v_{f}=v_{o}+at$ $v_{f}=19+3(4)=31$ $15=31+a(2)$ $a=\frac{15-31}{2}=-8$ $v_{f}^{2}=v_{o}^{2}+2 a \Delta x$ $\left(15\right)^{2} = \left(31\right )^{2}+2 \left(-8\right) \Delta x$ $2\left(-8\right) \Delta x = 225-961 = 736$ $\Delta x = 46$ 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
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Quote:
 Originally Posted by Chemist116 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? $\begin{array}{ll} 1.&\textrm{128 m}\\ 2.&\textrm{130 m}\\ 3.&\textrm{144 m}\\ 4.&\textrm{124 m}\\ 5.&\textrm{152 m} \end{array}$
Assuming the particle starts at $x(0) = 0$ ...

$\displaystyle \frac{dv}{dt} = 3$

$\displaystyle v(t) = 3t + C$

$v(6) = 15 \implies C = -3$

$\displaystyle \frac{dx}{dt} = 3t-3$

$x(t) = \dfrac{3t^2}{2} - 3t + C$

$x(4) = 100 \implies C = 88$

$x(t) = \dfrac{3t^2}{2} - 3t + 88$

$x(6) = 124$ October 21st, 2019, 08:57 PM   #3
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Quote:
 Originally Posted by skeeter Assuming the particle starts at $x(0) = 0$ ... $\displaystyle \frac{dv}{dt} = 3$ $\displaystyle v(t) = 3t + C$ $v(6) = 15 \implies C = -3$ $\displaystyle \frac{dx}{dt} = 3t-3$ $x(t) = \dfrac{3t^2}{2} - 3t + C$ $x(4) = 100 \implies C = 88$ $x(t) = \dfrac{3t^2}{2} - 3t + 88$ $x(6) = 124$
I was hoping that it could exist a pre-calculus approach to this, but this approach is fine. (Although I really appreciate that an alternate version using less advanced maths could be offered too). When I attempted to solve this as shown I overlooked that when you take the integral there is a constant.  October 22nd, 2019, 05:36 AM   #4
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Quote:
 Originally Posted by Chemist116 I was hoping that it could exist a pre-calculus approach to this, but this approach is fine. (Although I really appreciate that an alternate version using less advanced maths could be offered too). When I attempted to solve this as shown I overlooked that when you take the integral there is a constant. $v_f=v_0+at$

$v(6)=v(4)+3(2) \implies v(4)= 9$

$x(6) = x(4) + v(4) \cdot 2 + \dfrac{1}{2}(3) \cdot 2^2$

$x(6) = 100 + 18 + 6$ Tags displacement, find, instant, particle 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 Albertjhons088 Algebra 1 September 21st, 2015 07:10 AM jaredbeach Calculus 2 February 22nd, 2012 04:05 AM skipjack Elementary Math 6 January 19th, 2012 08:05 AM remeday86 Calculus 2 March 7th, 2009 07:24 PM

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