PHY 201
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An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.
• At what average rate is the automobile's acceleration changing with respect to the slope of the incline?
answer/question/discussion: ->->->->->->->->->->->-> :
First, I am going to get the acceleration for the two separate situations:
The first situation, you have the displacement 10 m, the time interval 8 s, and the initial velocity 0 m/s. So to start with, you can find the average velocity from the displacement and the time interval: 10 m / 8 s = 1.25 m/s. Then, if you know the average velocity and the initial velocity, you can get the final velocity: 1.25 m/s * 2 = 2.5 m/s. Next, you can get the change in velocities by subtracting: 2.5 m/s – 0 m/s = 2.5 m/s. Finally, if you know the change in velocity and the time interval you can get acceleration by: 2.5 m/s / 8 s = .3125 m/s^2.
The second situation can be solved just like the first, with just different variables: vAve = 10 m / 5 s = 2 m/s , 2 m/s * 2 = 4 m/s , 4 m/s – 0 m/s = 4 m/s , 4 m/s / 5 s = .8 m/s^2.
The change in the slope is: .10 - .05 = .05, and the change in acceleration is: .8 m/s^2 - .3125 m/s^2 = .4875 m/s^2. So, the average rate the automobile’s accelerations is changing with respect to the slope of the incline: .4875 m/s^2 / .05 = 9.75 m/s^2
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30 minutes
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