course Phy 231
5/12/2010 1:17
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.
Acceration= `dv/`dt, so (27m/s-8m/s)/13s= 1.5m/s^2
vAve= 27+8/2= 17.5
17.5m/s=’ds/13s so 13*17.5= 227.5cm
The slope of the graph represents the constant acceleration of the objects. Since acceleration is constant then the object is increasing at a constant rate. The area of a trapezoid is average height * width. The height is the run of the graph or the time and the rise of the slope of the graph is the velocity. So the area is 27+8/2= 17.5* 13=227.5 which is the distance traveled
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&#This looks very good. Let me know if you have any questions. &#