Hello
Let's solve the problem in the three different steps
1) Uniformly accelerated motion, with acceleration [tex]a_1 = 2.65~m/s^2[/tex] and for a total time of [tex]t_1=17~s[/tex]. The body is initially at rest, so the distance covered is given by
[tex]S= \frac{1}{2}a_1t_1^2=382.9~m [/tex]
Calling [tex]v_f[/tex] and [tex]v_i[/tex] the final and initial velocity, and since the [tex]v_i=0~m/s[/tex] because the body starts from rest, we can use
[tex]a= \frac{v_f-v_i}{t} [/tex]
to find the final velocity after this first leg:
[tex]v_{f}=v_i+a_1t_1=45~m/s[/tex]
And the average velocity in this first leg is
[tex]v_1= \frac{v_f+v_i}{2}=22.5~m/s [/tex]
2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: [tex]v_2=45~m/s[/tex]. This is also the average velocity of the second leg.
The total time of this second leg is [tex]t_2=1.60~min = 96~s[/tex]. The distance covered is given by
[tex]S_2=v_2t_2=45~m/s \cdot 96~s=4320~m[/tex]
3) Uniformly decelerated motion, with constant deceleration of [tex]a_3=-9.39~m/s^2[/tex] and for a total time of [tex]t_3=4.8~s[/tex]. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. [tex]v_i=45~m/s[/tex]. Therefore, the distance covered in this leg is given by
[tex]S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m[/tex]
The final velocity in this leg is given by
[tex]v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s[/tex]
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
[tex]v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s [/tex]
4) Finally, the total distance covered in the motion is
[tex]S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m[/tex]
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
[tex]v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s [/tex]
