Phy 231
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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: ->->->->->->->->->->->-> :
Since the ball is accelerating at 8 cm/s2 that means that every second the velocity is raising by 8. SO by starting at the 12 cm/s and adding 8 times 3 will be 36 cm/s.
. Assuming that acceleration is constant, what will be its average velocity during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The way you get average velocity is by taking the final velocity minus the starting velocity and dividing that by the time difference. (36-12)/(3-1) = 12m/s
. How far will it travel during this interval?
answer/question/discussion: ->->->->->->->->->->->-> :
The way to Find the displacement of the ball will be D=x2-x1 but first we have to find the distances traveled in the time. So for the first position since we start at time 0 we know that the distance will be 0. but the last position has a velocity and time so we find that by x2= V(T)^2 so x2= 36(3)^2 = 324 cm. So the ball has traveled by 324 cm.
I'm not sure where you got the equation x2= V(T)^2, but the equation is dimensionally inconsistent and is therefore cannot be valid.
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30 minutes
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Solution
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