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October 21st, 2019, 06:33 PM   #1
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Question 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?.
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October 21st, 2019, 08:15 PM   #2
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Quote:
Originally Posted by Chemist116 View Post
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 $x(0) = 0$ ...

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

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

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

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

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

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

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

$x(6) = 124$
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October 21st, 2019, 08:57 PM   #3
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Thumbs up Gee How could I overlooked integrals

Quote:
Originally Posted by skeeter View Post
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.
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October 22nd, 2019, 05:36 AM   #4
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Quote:
Originally Posted by Chemist116 View Post
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.
$v_f=v_0+at$

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

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

$x(6) = 100 + 18 + 6$
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