A body moves along one dimension with a constant acceleration of 4.35 m/s2 over a time interval. At the end of this interval it has reached a velocity of 13.8 m/s.(a)If its original velocity is 6.90 m/s, what is its displacement (in m) during the time interval? m (b)What is the distance it travels (in m) during this interval? m(c) A second body moves in one dimension, also with a constant acceleration of 4.35 m/s2 , but over some different time interval. Like the first body, its velocity at the end of the interval is 13.8 m/s, but its initial velocity is −6.90 m/s. What is the displacement (in m) of the second body over this interval? m (d) What is the total distance traveled (in m) by the second body in part (c), during the interval in part (c)?

[tex]\begin{gathered} S=v_ot+\frac{1}{2}at^2 \\ =6.9\times1.59+\frac{1}{2}\times4.35\times(1.59)^2 \\ =10.971+5.498 \\ =16.469\text{ m} \end{gathered}[/tex]

(b) The initial speed will be equal to the initial velocity.

Thus, the distance will be 16.469 m.

(c) The initial velocity of the second object is

[tex]v^{\prime}_o=\text{ -6.9 m/s}[/tex]

The final velocity of the second object is

[tex]v^{\prime}_f=13.8\text{ m/s}[/tex]

The acceleration of the second object is

[tex]a^{\prime}=\text{ 4.35 m/s}[/tex]

The time taken will be

[tex]\begin{gathered} t^{\prime}=\frac{(v_f-v_o)}{a} \\ =\frac{13.8-(-6.9)_{}}{4.35} \\ =4.75\text{ s} \end{gathered}[/tex]

We have to find the displacement of the second object.

Let the displacement of the second object be denoted by S'.

The displacement of the second object will be

[tex]\begin{gathered} S^{\prime}=v^{\prime}_ot^{\prime}+\frac{1}{2}a^{\prime}(t^{\prime})^2 \\ =-6.9\times4.75+\frac{1}{2}\times4.35\times(4.75)^2 \\ =-32.775+49.073 \\ =16.298\text{ m} \end{gathered}[/tex]

(d) The initial speed of the object will be

[tex]v^{\doubleprime}_o=6.9\text{ m/s}[/tex]

We have to find the distance.

Let the distance be denoted by d'.

The distance of the second object will be

[tex]\begin{gathered} d^{\prime}=v^{\doubleprime}_ot^{\prime}+\frac{1}{2}a^{\prime}(t^{\prime})^2 \\ =6.9\times4.75+\frac{1}{2}\times4.35\times(4.75)^2_{} \\ =32.775+49.073 \\ =81.848\text{ m} \end{gathered}[/tex]