Weighted Average of the VelocitiesDate: 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/ |
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