Converting Acceleration to Velocity

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SUMMARY

The discussion clarifies that velocity is not equivalent to the square root of acceleration. Instead, acceleration (A) is defined as the rate of change of velocity (V) over time (t), expressed as A = V/t. The correct relationship between distance (d), acceleration, and time is given by the kinematic equation d = Vi*t + 0.5*a*t^2. A user calculated the time taken for a 100g mass to travel 10 cm under a constant force of 0.0000000132944 N, resulting in approximately 1226.54 seconds, confirming the proper application of the kinematic equation.

PREREQUISITES
  • Understanding of basic kinematics
  • Familiarity with Newton's second law of motion
  • Knowledge of units of measurement (meters, seconds, Newtons)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the kinematic equations in detail, particularly d = Vi*t + 0.5*a*t^2
  • Learn about Newton's second law and its implications for acceleration
  • Explore resources on significant figures and their importance in physics calculations
  • Review introductory physics textbooks or online courses focusing on motion and forces
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Students studying physics, educators teaching kinematics, and anyone seeking to understand the relationship between acceleration, velocity, and distance in motion.

Thrawn
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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.
 
Last edited:
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Thrawn said:
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.
 
Precisely, acceleration is the derivative 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
 
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?
 
Don't you have a textbook that defines acceleration?
 
This site isn't bad: http://www.physicsclassroom.com/Class/1DKin/U1L6a.html"

And check out our own tutorial section: https://www.physicsforums.com/showthread.php?t=95426"
 
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HallsofIvy said:
Don't you have a textbook 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...
 
Doc Al said:
This site isn't bad: http://www.physicsclassroom.com/Class/1DKin/U1L6a.html"

And check out our own tutorial section: https://www.physicsforums.com/showthread.php?t=95426"


Thank you.
 
Last edited by a moderator:
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
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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