course phy201
6/19 5
The velocity of an object increases at a uniform rate from 5 m/s to 27 m/s in 14 seconds.How far does it travel and what is its acceleration?
Sketch the corresponding velocity vs. clock time graph.
Explain the meaning of the slope of the graph and its area.
In order to calculate the displacement of the object we first need to know the vAve and the change in clock time. By adding 5 m/s and 27m/s and then dividing by 2 we get an average velocity of 16m/s.
16m/s * 14s( time interval)= 224 m for distance traveled.
Now to calculate acceleration we take the change in velocity divided by the change in time.
22m/s/ 14 s= 1.6 m/s/s
when you graph this same problem the diagonal becomes part of a right triange with sides measuring 22 for the change in velocity(rise) and 14 for the change in time(run).
the slope of this line is then 22/14= 1.6 which is also the acceleration as stated above. To calculate the area you take the ave height, which in this case is also the average velocity since acceleration is constant, and multiply that by the width which is also the change in time.
16*14= 224m which as you can see is the same answer for the distance traveled.
"
&#This looks very good. Let me know if you have any questions. &#
#$&*