Weighted Average of the Velocities

Date: 01/30/2003 at 03:46:16
From: Garin 
Subject: Average velocity

A truck on a straight road starts from rest and accelerates at 2.0 
m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 
20 s at constant speed until the brakes are applied, stopping the 
truck in a uniform manner in an additional 5.0 s. 

(a) How long is the truck in motion? 
(b) What is the average velocity of the truck for the motion 
    described?

Date: 01/30/2003 at 11:04:26
From: Doctor Edwin
Subject: Re: Average velocity

Hi, Garin.

The trick here is to get the average velocity for each of the three 
phases of the trip, then do a weighted average of the three 
velocities.

Average velocity for phase 2 is easy, 20 m/s. During phases 1 and 3, 
the velocity changes linearly. That means that we can just average 
the starting and ending velocities to get the velocity for the entire 
phase.

Once you've got those, you'll do a weighted average with respect to 
the time spent in each phase:

           (v_1 * t_1) + (v_2 * t_2) + (v_3 * t_3)
  v_avg =  ---------------------------------------
                           t_total

Just for fun, I'll point out that the same set of equations has 
another interpretation that works just as well. Average velocity is 
just distance over time, right? So if we figure out how far the truck 
went in each phase, add them all up, and divide by the total time, 
you'll also get the same answer:


                d_1     +    d_2     +      d3
  v_avg =  ---------------------------------------
                           t_total

But the middle terms in those two equations are the same:

    d_2 = v_2 * t_2

and so are the first terms:

          a(t_1)^2
    d_1 = --------
              2

           v
     a  =  -
           t

          v_1_end * (t_1)^2
    d_1 = -----------------
                2(t_1)


            v_1_end * t_1
    d_1 =   -------------
                  2
    
But since the starting velocity is zero, we can add it in wherever we 
want:

           (v_1_start + v_1_end)
    d_1 =  ---------------------- * t_1
                      2

which means that our first terms are the same as well, and so are our 
third terms.

Anyway, there you have it. Total distance traveled divided by time, 
or a weighted average of velocities with respect to time, it works 
out to the same thing.

- Doctor Edwin, The Math Forum
  http://mathforum.org/dr.math/