#$&*
course Phy 121
9/10 6:27 pm
If an object increases velocity at a uniform rate from 5 m/s to 21 m/s in 12 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 5 m/s and whose velocity 12 seconds later is 21 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
acceleration = `dv/`dt
a = (21m/sec - 5m/sec)/12 sec
a = (16m/sec) /12 sec = approximately 1.3m/s^2
For the distance traveled:
`ds = `vAve * dt
`ds = (5m/s+21m/s)/2 * 12 seconds
`ds = (26m/s)/2 *12 seconds
`ds = 13m/s * 12 seconds = 156 meters traveled during the interval.
The slope of the graph by definition is the rise / run. In this case the rise is the `dv and the run is the `dt. Therefore the slope is the `dv/`dt which is the acceleration.
The area beneath the graph represents the change in position. The average altitude which in this case is vAve is multiplied by the width which in this case the change in clock time gives you the change in position. It’s a graphical representation of `ds = `vAve * dt.
"
&#This looks very good. Let me know if you have any questions. &#