course Phy 231
If an object increases velocity at a uniform rate from 6 m/s to 20 m/s in 13 seconds, what is its acceleration and how far does it travel?Given values are v0 = 6m/s, vf = 20m/s, and `dt = 13s. The goal is to find acceleration and the change in position.
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In order to find position, vAve and `dt are needed. `dt is given, and vAve can be found using v0 and vf:
vAve = (vf + v0)/2
vAve = (20m/s + 6m/s)/2 = 26m/s/2 = 13m/s
`ds = vAve * `dt
`ds = 13m/s * 13s
`ds = 169m
In order to find the acceleration, `dv and `dt are needed. `dt was given and `dv can be found using v0 and vf:
`dv = vf - v0
`dv = 20m/s - 6m/s = 14m/s
a = `dv/`dt
a = 14m/s/13s = 1.1m/s^2
Sketch a velocity vs. clock time graph for an object whose initial velocity is 6 m/s and whose velocity 13 seconds later is 20 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
The graph would look have the vertical axis be velocity. The horizontal axis would be time. There would be a point at (0,6) and a point at (13,20). There would be a straight line connecting these points. The slope of this graph is the average acceleration because the rise is the change in velocity and the run is the change in clock time. The area under the graph is the change in position because the area of the trapezoid is calculated just as position would be calculated from the average velocity * the clock time."
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