0
$\begingroup$

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)

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

$v_{f}=19+3(4)=31$

$15=31+a(2)$

$a=\frac{15-31}{2}=-8$

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

$\left(15\right)^{2}=\left(31\right)^{2}+2(-8)\Delta x$

$2(-8)\Delta x = 225-961 = 736$

$\Delta x = 46$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

If it has a constant acceleration of $3\frac{m}{s}$ and reaches a speed of $15\frac{m}{s}$ after $6$ seconds, then its initial speed $v_0$ at four seconds was $9\frac{m}{s}$.

To find the displacement, use the equation $\Delta x= v_0t +\frac{1}{2}at^2.$ And so its displacement at $6$ seconds should be $100+(9(2)+\frac{1}{2}\cdot3\cdot 2^2)=124$ m.

$\endgroup$
10
  • $\begingroup$ But how do I exactly get to that answer?. This is why I shown the steps I attempted using the position equation. Why I can't obtain that speed from there?. $\endgroup$ Oct 22, 2019 at 1:36
  • $\begingroup$ I did noted your edit but it still doesn't address my problem which is exactly on what part of my assumptions did I made a mistake. $\endgroup$ Oct 22, 2019 at 1:37
  • $\begingroup$ In other words why I cannot obtain the initial speed $v_{o}=9$ from the position equation. $\endgroup$ Oct 22, 2019 at 1:38
  • $\begingroup$ Then what should had been changed there to get to $9$ or it cannot be found given that information? $\endgroup$ Oct 22, 2019 at 1:44
  • $\begingroup$ Okay If I use the conditions you mention then what. I mean $x(6)=?$ ? $\endgroup$ Oct 22, 2019 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.