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.
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.