course phys 201
Over a period of 6 seconds, an object increases its velocity at a uniform rate from 9 m/s to 23 m/s •What is its acceleration and how far does it travel?
Acceleration= change in velocity/time= 23-9/6=2.333m/s^2
Distance travelled= xo+vot+1/2at^2= 0+(9)t+1/2(2.333)t^2=95.994m
good use of the equation, but you should also understand that `dx = vAve * `dt = (9 m/s + 23 m/s) / 2 * 6 sec = 96 m.
• Graph velocity vs. clock time for this object and explain what the slope of the graph means and why, and also what the area means and why.
The slope denotes the average velocity because it is a function of velocity as time progresses. The area represents the total displacement because the rectangle at the bottom has an area of base. Height (v0t) while the rectangle has an area of ½ base. Height = ½ t. (v-vo). V- Vo= at hence the combination gives v0t + 1/2.at^2 which is the formula for s(distance/displacement)
very good.
&#This looks good. See my notes. Let me know if you have any questions. &#