Answer:
The final position of the particle will be at x = -1.0 meters for t = 6.0 seconds.
Explanation:
The displacement can be found as the integral or area of the velocity function given between 2 values of the time.
[tex]\displaystyle \Delta s_x = \int\limits^a_b {V_x(t)} \, dt[/tex]
Also the displacement is one dimension is defined as the difference between 2 positions, that is
[tex]\Delta s_x = x(t_f) -x(t_i)[/tex]
So for the exercise we have
[tex]\Delta s_x =x(6)-x(1)[/tex]
And we know that x(1) = 2.0, so we can write
[tex]x(6) = x(1) +\Delta s_x \\x(6) = 2.0 m + \Delta s_x[/tex]
Thus if we find the areas after t = 1.0 seconds up to 6.0 seconds, we can just add them to 2.0 meters to find the position at t = 6.0 seconds.
Finding Areas after t = 1.0 s
After t = 1.0 seconds, we have 2 triangles, one that is above the horizontal axis, that is between t = 1.0 to t = 2.0 seconds, and we have one triangle below the horizontal axis, that is between t = 2.0 seconds to t = 6.0 seconds.
For the first area we have:
- Base, b = 2.0 -1.0 = 1
- Height, h = 2.0-0= 2
Thus the area of the triangle is
[tex]A_1 = \cfrac 12 bh \\ A_1 = \cfrac 12 (1)(2)\\A_1 = 1 m[/tex]
The first area from 1 to 2 seconds is 1 meter value.
For the second area we have:
- Base, b = 6.0-2.0 = 4
- Height, h = 0-(-2.0) = 2
Notice that the height is measured from the bottom of the triangle to the horizontal line.
Thus the area of the triangle is
[tex]A_2 = \cfrac 12 bh \\ A_2 = \cfrac 12 (4)(2)\\A_2 = 4 m[/tex]
The first area from 2 to 6 seconds is 4 meter value, however it is below the axis, so we consider a signed area, that means it is negative if it is below the horizontal axis.
In Physic terms, having a negative area or a negative velocity under the horizontal line means that the object is moving to the left.
The displacement now is the sum of all areas, if the areas are above the horizontal axis, they are positive, if they are below the horizontal axis, they have a negative in front for the signed area.
[tex]\Delta s_x =A_1 -A_2 \\\Delta s_x =1-4\\\Delta s_x =-3[/tex]
So the position at t =6.0 seconds will be
[tex]x(6) = 2.0 m + \Delta s_x\\x(6) = 2.0 m + (-3 m)\\\boxed{x(6) = -1.0 m}[/tex]
Thus the final position is at -1.0 meters since the particle was moving to the left after t = 2 seconds.
