#$&*
Phy 231
Your 'cq_1_05.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
The problem:
A ball accelerates at 8 cm/s^2 for 3 seconds, starting with velocity 12 cm/s.
• What will be its velocity after the 3 seconds has elapsed?
answer/question/discussion: ->->->->->->->->->->->-> :
8cm/sec/sec * 3sec= 24cm/sec
12cm/sec + 24cm/sec= 36cm/sec
#$&*
• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= (36cm/sec +12cm/sec)/2= 24cm/sec
#$&*
• How far will it travel during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
ds= vAve * dt= 24cm/sec * 3sec= 72cm
#$&*
** **
15 min
** **
@& Very good.
There are a number of important connections here to the calculus of the situation:
Note that if the interval begins at t = 0, the velocity function is 12 cm/s + 8 cm/s^2 * t.
Integrating from t = 0 to t = 3 s we get antiderivative 12 cm/s * t + 8 cm/s^2 * t^2 / 2. The antiderivative changes from 0 cm to 12 cm/s * 3 s + 8 cm/s^2 * (3 sec)^2 / 2 = 72 cm, a change of 72 cm - 0 cm = 72 cm.
The antiderivative function 12 cm/s * t + 8 cm/s^2 * t^2 / 2 is the position function, provided we assume position 0 cm at clock time 0 sec.
The average value of a function over an interval is equal to its integral divided by the length of the interval, in this case 72 cm / (3 sec) = 24 cm/s. This reconciles with the value we obtain from the initial and final velocities, and the linearity of the velocity vs. clock time function.
*@