Respuesta :
We can solve this by using the equations of kinematics below
[tex]\begin{gathered} v=u+at \\ \Delta x=ut+\frac{1}{2}at^2 \\ v^2=u^2+2a\Delta x \end{gathered}[/tex](a).
[tex]\begin{gathered} a=3.95m/s^2 \\ v=11.4\text{ m/s} \\ u=5.70\text{ m/s} \\ \Delta x=\text{?} \\ v^2=u^2+2a\Delta x \\ 11.4^2=5.70^2+2\times3.95\times\Delta x \\ 129.96-32.49=7.9\Delta x \\ \Delta x=\frac{97.47}{7.9} \\ \Delta x=12.3379746835\approx12.34m \end{gathered}[/tex](b)
Since the object moves in a straight part the distance will also be 12.34 m.
(c)
[tex]\begin{gathered} u=-5.70 \\ v=11.4\text{ m/s} \\ a=3.95 \\ v^2=u^2+2a\Delta x \\ 11.4^2=-5.70^2+2\times3.95\times\Delta x \\ \Delta x=12.34\text{ m} \end{gathered}[/tex](d)
The total distance is just the total distance covered without direction. In part c the initial velocity was negative, this means it traveled in the opposite direction therefore that part distance can be calculated below
[tex]\begin{gathered} u=-5.70\text{ m/s} \\ v=0\text{ m/s} \\ a=3.95m/s^2 \\ v^2=u^2+2a\Delta x \\ (0)^2=-5.70^2+2\times3.95\times\Delta x \\ -32.49=7.9\Delta x \\ \Delta x=-\frac{32.49}{7.9} \\ \Delta x=-4.11265822785\approx-4.11\text{ m} \end{gathered}[/tex]Then the path where the initial velocity was zero and the final velocity was 11.4 m/s can be calculated as follows
[tex]\begin{gathered} u=0\text{ m/s} \\ v=11.4\text{ m/s} \\ a=3.95m/s^2 \\ v^2=u^2+2a\Delta x \\ 11.4^2=0^2+2\times3.95\times\Delta x \\ 129.96=7.9\Delta x \\ \Delta x=\frac{129.96}{7.9} \\ \Delta x=16.4506329114\approx16.45\text{ m} \end{gathered}[/tex]Therefore,
total distance = 4.11 m + 16.45 m = 20.56 m
