course Phy 201
Assignment 4An 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?
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Change in distance / Time elapsed = average velocity
From start to point A = 75 meters – 0 meters / 4 seconds – 0 seconds = 75 m/ 4 sec = 18.75 m/s
From point A to point B = 115 m – 75 m / 11 s – 4 s = 40 m / 7 s = 5.7 m/s
From start to point B = 115 m – 0 m / 11 sec / 0 sec = 115 m / 11 s = 10.5 m/s
The problem does not state that where the car is or what its velocity is at t = 0. Only the conditions at the beginning and end of the given interval are specified, and nothing else can be assumed.
Change in velocity / Time elapsed = average acceleration
From start to point A=10 m/s – 0 m/s / 4 s = 2.5 m/s^2
From point A to point B=20.5 m/s – 10 m/s / 7 sec = 1.5 m/s^2
From start to point B= 20.5 m/s – 0 m/s / 11 sec = 1.86 m/s^2
The acceleration was not uniform because the speed of the automobile was not traveling at a constant rate.
Your solution would be correct given your assumptions about velocity and position at t = 0. However this information was not included and cannot be used.
Please submit a revision that doesn't rely on these assumptions.
In relation to the last question, note the following:
&#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? &#