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Old Feb 27th 2011, 07:47 AM   #1
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finding initial velocity simple kinematics question.

A bike rider accelerates constantly to a velocity of 7.5m/s during 4.5s. The bikes displacement is 19m. What was the initial velocity of the bike?

d= (v1+v2)/2 t

19m=(v1+7.5m/s) 4.5s

now this next step I've forgotten how to do I'm 2 years out of school so thats probably why. I have the answers for these questions but im unsure how it got to this point.

v1=2d/t-v2

=2(19m)/(4.5s) - 7.5m/s
=0.94m/s
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Old Feb 27th 2011, 10:25 AM   #2
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Originally Posted by davie08 View Post
d= (v1+v2)/2 t

19m=(v1+7.5m/s) 4.5s
This last line is wrong. You left out the 2.

Originally Posted by davie08 View Post
v1=2d/t-v2

=2(19m)/(4.5s) - 7.5m/s
=0.94m/s
However these are correct.

-Dan
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Old Feb 27th 2011, 12:54 PM   #3
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okay but the second part I dont understand how it got to that point from

d=(v1+v2)/2 (t)

to

v1=2d/t-v2
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Old Feb 27th 2011, 01:25 PM   #4
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Originally Posted by davie08 View Post
okay but the second part I dont understand how it got to that point from

d=(v1+v2)/2 (t)

to

v1=2d/t-v2
d = (v1 + v2)/2 (t)


You want to isolate the v1. Start with dividing both sides by t:
d/t = (v1 + v2)/2

Multiply both sides by 2:
2d/t = v1 + v2

Now subtract v2 from both sides:
2d/t - v2 = v1

-Dan
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