course PHY 152
The problem:A graph is constructed representing velocity vs. clock time for the interval between clock times t = 5 seconds and t = 13 seconds. The graph consists of a straight line from the point (5 sec, 16 cm/s) to the point (13 sec, 40 cm/s).
• What is the clock time at the midpoint of this interval?
**9 seconds (13-5=8/2=4+5=9)
answer/question/discussion:
• What is the velocity at the midpoint of this interval?
**28cm / 9s = 3.1cm/s (40-16=24/2=12+16=28)
answer/question/discussion:
• How far do you think the object travels during this interval?
**28cm-16cm=12cm
### I should have multiplied 28cm by 8s to = 224cm
answer/question/discussion:
• By how much does the clock time change during this interval?
**9s-5s=4s
###this should have been 13s-5s=8s
answer/question/discussion:
• By how much does velocity change during this interval?
**12cm*2=24cm
answer/question/discussion:
• What is the average rate of change of velocity with respect to clock time on this interval?
** 4.8cm/s * 1s=4.8cm/s^2
###This should have been 24cm/s / 8s=3cm/s
3cm/s * 1s= 3cm/s^2
answer/question/discussion:
• What is the rise of the graph between these points?
??Which points is this asking for?
###The rise was the 40cm/s-16cm/s=24cm/s
answer/question/discussion:
• What is the run of the graph between these points?
??
###The run was 13s-5s=8s
answer/question/discussion:
• What is the slope of the graph between these points?
??
###The slope would then be 24cm/s / 8s=3cm/s*1s=3 cm/s^2
answer/question/discussion:
• What does the slope of the graph tell you about the motion of the object during this interval?
??
###It would show an increasing linear line representing the calculations from above
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&#Good responses. Let me know if you have questions. &#