Question

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2. Maintain a constant velocity for the next 1.60 min. 3. Apply a constant negative acceleration of −9.39 m/s2 for 4.80 s.
(a) What was the total displacement for the trip?

b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip

Answer 1

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$. The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$. This is also the average velocity of the second leg. 
The total time of this second leg is $t_2=1.60~min = 96~s$. The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$. Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
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
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$