Converting Acceleration to Velocity

  • Thread starter Thrawn
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  • #1
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Homework Statement


Is velocity equivalent to the square root of accelleration?


Homework Equations


V = d/t


The Attempt at a Solution



A = accelleration, expressed in meters per second squared

V = distance (m) divided by time (s)

So V = sqrt A ?



In the context of the question: How long does it take an object with a mass of 100g (.1kg) mass to travel 10 cm (.1m), with a force (assumed to be constant) of 0.0000000132944 N.
 
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Answers and Replies

  • #2
Doc Al
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Is velocity equivalent to the square root of accelleration?
Uh... no.

A = accelleration, expressed in meters per second squared

V = distance (m) divided by time (s)

So V = sqrt A ?
If velocity were the square root of acceleration, it would have units of square root of distance divided by time.

In the context of the question: How long does it take an object with a mass of 100g (.1kg) mass to travel 10 cm (.1m), with a force (assumed to be constant) of 0.0000000132944 N.
Time for you to review basic kinematics. Find an expression that relates distance with time for a given acceleration.
 
  • #3
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Precisely, acceleration is the derivitive of velocity, expressed in m/s per second. Its the rate of change of velocity, or V/t. That means V = At

Your question relates distance (not velocity) to acceleration, in which case you need to find the constant acceleration and then plug it into the kinematic equation D = vo + .5at^2
 
  • #4
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If possible, could someone direct me to a website which provides an introduction (prefferably oriented towards beginners with very little prior knowledge of the subject or complementary subjects) to the kinematics involved in this equation?
 
  • #5
HallsofIvy
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Don't you have a text book that defines acceleration?
 
  • #7
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Don't you have a text book that defines acceleration?
I wish I had a textbook for this... but alas, it is not covered in the Ontario Grade 8 curriculum, as such a physics textbook would be rather unneccessary, as it is not taught...
 
  • #9
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Having found the appropriate kinematic equation, I have calculated the following:

d = Vi*t + 0.5*a*t^2

= 0m/s*t + 0.5*a*t^2

= a*t^2/2

2d= a*t^2

2(0.1m)/a = t^2

0.2/0.000000132944m/s^2 = t^2

sqrt 1 504 392.827055000617 = t

t = 1 226.5369244 s

Is the value for t, one thousand two hundred twenty - six seconds correct, and was the equation used correctly?
 
  • #10
Doc Al
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That looks fine. A few comments:

Express numbers such as the force using exponential notation (like 1.32944 × 10^-8 N); they are much easier to read that way instead of a string of zeroes.

Take care with significant figures. It looks like your values for mass and distance only have one significant figure--yet your answer has around 20. Not very realistic! (Just because the calculator carries all those digits doesn't mean you should keep them in your answer.)
 

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