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:


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:



$100=x(4)=4v_{ox}+ \frac{3\times 4 \times 4}{2}$

$100=4v_{ox}+ \frac{3\times 4 \times 4}{2}$

$25=v_{ox}+ 6$


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)





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

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

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

$\Delta x = 46$

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.


1 Answer 1


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.

  • $\begingroup$ 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

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