# How do I find the displacement of a particle at a given instant when the initial speed is not given?

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.
• In other words why I cannot obtain the initial speed $v_{o}=9$ from the position equation. Oct 22, 2019 at 1:38
• Then what should had been changed there to get to $9$ or it cannot be found given that information? Oct 22, 2019 at 1:44
• Okay If I use the conditions you mention then what. I mean $x(6)=?$ ? Oct 22, 2019 at 1:48