course phy 201
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Week 2 Quiz 1 Version 1If 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
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.
The acceleration is found by a = ‘dv/’dt
= (27m/s -8m/s)/13s
a = 1.5m/s
I have a graph but it won't copy in!!!!!!
Describe graphs. Even if you make a graph, it's essential to describe it to show that you understand what it is telling you.
The slope of the line is how you can find the acceleration. If you take two points on the line and then make a triangle with them. You can then obtain the slope which equal rise/run. The change in velocity divided by the change in time.
When you find this displacement of the you multiply the change in velocity by the change in time which really equals the area under the slope. Because the change in velocity is the altitude and the change in time if the base.
Good answer on acceleration.
However, you need to be sure you are distinguishing change in velocity from average velocity.
Change in velocity is the rise from one point to the other, on a graph the change in altitude, and is not the same thing as average velocity.
The average velocity is identified the average altitude, which is very different on most graphs from the change in altitude. Average velocity and change in velocity are very different quantities.
Let me know if you have questions.