*Image by Columbia Center for Urban Agriculture*

Taken by Li Tang

*Question by Icarus52*: Can you help me with Average Velocity and Acceleration?

2.If you take the average velocity over some period of time (say 1 hour on a car trip), is it possible that the instantaneous velocity at any moment during that period is larger than the average velocity?

4.Given a plot of distance versus (time)^2 , what feature of the curve represents acceleration? (The slope?, the y-axis intercept?, 2 times the intercept?, …) Please explain your answer, briefly.

**Can you help? Leave your own answer in the comments!**

2. Not only possible, but very likely. To illustrate: Assume you travel the distance S = Vavg T to grandma’s house for her good southern fried chicken. She’ll serve in T = 2 hrs and you live S = 60 mi away in a neighboring city. Due to speed limits, which you follow as a law abiding citizen, you must travel no more that 25 mph over 20 mi urban streets before being able to speed up on the final 40 mi to grandma’s place in the country.

What speed must you average to get the her place on time? Vavg = S/T = 60/2 = 30 mph. How fast must you drive over the last 40 mi to get to that chicken dinner on time given that you cannot go above 25 mph for 20 mi?

We have T = s1/v1 + s2/v2 = t1 + t2 = 2 hrs = 20/25 + 40/v2 where v2 is the current speed you must drive over the rural roads to get the drumstick. So 2 – 4/5 = 40/v2 and v2 = 40/(6/5) = 200/6 = 33.33333333 = 33.3 mph > 30 mph = Vavg QED.

Bottom line. Because of the nature of an average, there will always…always…be some points in that average that are greater than the average. And of course there will be values lower (like the 25 mph) than the average as well.

2. S = UT + 1/2 AT^2; the slope of this curve dS/dT = V(T) = U + AT is the velocity at the time T. The Y intercept is S(0) = 0 when T = 0. That is, the curve begins at the origin p(0,0). No clue what you mean by 2 time the intercept…2 X 0 = 0 it appears to me.

oldprof

January 29, 2014 at 2:54 am