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Old Aug 14th 2012, 03:09 AM   #1
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Problem regarding Acceleration

1. An object has a constant acceleration of 8m/s/s. How far will the object reach after 5 seconds?
Given:
a=8 m/s^2
t= 5 s
d=?

Most of the class solution:
They first derived final velocity. The formula for a is a= (vf-vi)/(tf-ti)

8m/s/s=(vf-vi)/5
vf=40m/s

they derived distance by multiplying velocity and time. Therefore:
d=(40m/s)(5s)
d=200m

My solution:
Through advanced reading, I managed to learn a formula d=1/2(at^2)

d=1/2[(8 m/s/s)(5^2)]
d=1/2[(8m/s/s)(25)]
d=1/2(200)
d=100m

Which one is correct? My solution or my class' solution?
2. A body has a constant acceleration of 6m/s/s and traveled 50m. What is its velocity?
Given:
a=6 m/s^2
d=50 m
v=?
t=?


d=(v^2)/2a
50=(v^2)/2(6)
50=v^2/12
600=v^2
v=24.49489743 m/s

t=2.041241452 s

Is my solution correct or does it need some work?

Last edited by RPSGCC733; Aug 22nd 2012 at 05:29 AM.
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Old Aug 14th 2012, 06:53 AM   #2
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For 1 you are correct. The formula d = (1/2) at^2 gives the distance travelled under constant acceleration with no initial velocity. Your klass mates have calculated the final velocity OK, but made the mistake of assuming that the object travels at that final velocity for the full five seconds, but of course it doesn't.

For 2 you are also correct - good work!
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Old Aug 14th 2012, 08:32 AM   #3
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Originally Posted by ChipB View Post
For 1 you are correct. The formula d = (1/2) at^2 gives the distance travelled under constant acceleration with no initial velocity. Your klass mates have calculated the final velocity OK, but made the mistake of assuming that the object travels at that final velocity for the full five seconds, but of course it doesn't.

For 2 you are also correct - good work!
For number 2, what type of velocity should I look for? The question did not state if it was final velocity or average velocity. I solved for the final velocity and time should be 4 seconds not 2. Does the term velocity refer to average?
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Old Aug 15th 2012, 04:42 AM   #4
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The problems states that the object "traveled" 5 meters (past tense) and asks what "is" its velocity (present tense), so it seems that they want its most recent velocity, hence final velocity. That's almost always what is asked for in these types of problems. If they want average velocity the problem would specifically ask for it.

And yes - the time for (2) is a bit over 4 seconds. I saw that you had final velocity correct and didn't notice that they also wanted time.

Last edited by ChipB; Aug 15th 2012 at 04:45 AM.
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