Since the object is moving at constant acceleration this is an uniformly accelerated motion and then we can use the following equations:
[tex]v_f^2-v_0^2=2a\Delta x[/tex]a)
In this case the initial velocity is 6.50 m/s, the final velocity is 13.0 m/s and the acceleration is 3.00 m/s², then we have:
[tex]\begin{gathered} \Delta x=\frac{(13)^2-(6.5)^2}{2(3)} \\ \Delta x=21.13 \end{gathered}[/tex]Therefore, the displacement during this time interval is 21.23 m
b)
Since the object was moving in the positive direction from the beginning the distance is equal to the displacement; therefore, the distance during this time interval is 21.23 m
c)
In this case the initial velocity is -6.50 m/s, the final velocity is 13.0 m/s and the acceleration is 3.00 m/s², then we have:
[tex]\begin{gathered} \Delta x=\frac{(13)^2-(-6.5)^2}{2(3)} \\ \Delta x=21.13 \end{gathered}[/tex]Therefore, the displacement during this time interval is 21.23 m
d)
In this case we need to calculate the distance in two parts; first we need to calculate how much the object moves while its moving to the left until it stops; for this part the initial velocity is -6.50 m/s and the final velocity is zero, then we have:
[tex]\begin{gathered} \Delta x=\frac{0^2-(-6.5)^2}{2(3)} \\ \Delta x=-7.04 \end{gathered}[/tex]Now we need to determine the displacement from when the velocity is zero until it reaches 13 m/s, then we have:
[tex]\begin{gathered} \Delta x=\frac{13^2-0^2}{2(3)} \\ \Delta x=28.17 \end{gathered}[/tex]Finally we add the absolute value of this displacements to calculate the distance:
[tex]d=7.04+28.17=35.21[/tex]Therefore, the distance the object travelled is 35.21 m
