Phy 121
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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: ->->->->->->->->->->->-> :
Acceleration is equaled to change in velocity divided by change in time. Therefore:
Acceleration= (final velocity-initial velocity)/(final time- initial time)
8 cm/sec^2=(final velocity- 12 cm/sec)/(3 sec)
Final Velocity= 36 cm/sec
• Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= `ds/`dt
To find the final and initial positions we have to multiply the velocities by the time interval:
36 cm/sec * 3 sec= 108 cm
12 cm/sec * 3 sec= 36 cm
vAve= (108 cm- 36 cm)/ 3 sec= 24 cm/sec
• How far will it travel during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
We can find the interval of distance by subtracting the initial position from the final position:
(108 cm- 36 cm)= 72 cm
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20 minutes
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&#Very good responses. Let me know if you have questions. &#