RPwk2quiz

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course Phys201

6/11 @9:15pm

If an object increases velocity at a uniform rate from 8 m/s to 27 m/s in 13 seconds, what is its acceleration and how far does it travel?Sketch a velocity vs. clock time graph for an object whose initial velocity is 8 m/s and whose velocity 13 seconds later is 27 m/s. Explain what the slope of the graph means and why, and also what the area means and why.

Rise/Run= `dv/`dt

Rise/Run=27-8/13

Acceleration=2 m/s

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You are not using units, and the units of your result are incorrect.

The procedure, and the numbers, are however correct.

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Displacement=avg velocity*time

Avg velocity=`ds/`dt

27-8=19/13

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There are no positions numerically equal to 27 and 8. So this calculation is not correct.

Note that when acceleration is uniform, the average velocity for an interval is the average of the initial and final velocities on that interval.

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2*13=26 m/s

The slope is calculated between 2 points on a graph (rise/run). With this we obtain the average velocity between corresponding clock times. The area under the velocity vs. clock time has slopes which give us back our average velocities.

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You're on the right track, but you have in a couple of places misidentified the given information.

And you need to use units.

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