assign4version1

course phy 201

An automobile traveling a straight line is at point A at clock time t = 4 sec, where it is traveling at 10 m/s, to point B at clock time t = 11 sec, where it is traveling at 20.5 m/s. Point A is 75 meters from the starting point and point B is 115 meters from the starting point. What are the average velocity and the average acceleration of the automobile during the specified time? What evidence is there that the acceleration is or is not uniform?

average velocity is teh change in position / the change in elapsed time

so 115m-75m / 11s-4s

40m/7s = 5.7m/s

average acceleration is the change in velocity / the change in elasped time so 20.5 m/s - 10m/s / 11s-4s

10.5m/s / 7s = 1.5 m/s/s

i would say the acceleration is not uniform since it takes 75 meters to go 10 m/s then only 40 more meters to travel 20.5 m/s that to me is proof that acceleration is of course taking place but not in a uniform manner. "

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Average velocity is defined as average rate of change of position, vAve = `ds / `dt.Average velocity is equal to the average of initial and final velocities, (vf + v0) / 2, if acceleration is uniform. If average velocity is not equal to the average of initial and final velocities, then acceleration is not uniform. This information can be used to determine whether or not acceleration is uniform. What is your conclusion? &#

&#You cannot assume that the position at t = 0 is 0. This is not a stated condition of this problem, and is not generally true. For example, when you use the TIMER program the clock time at the beginning of a timed interval is not 0.

However to make this calculation you have to make that assumption. Another way of saying this is that you don't learn anything by dividing position by time; the only quantity that has meaning is change in position / change in clock time. Just using a single given position and a single given clock time does not yield an average velocity, unless it is known that position is 0 at t = 0. &#