Distance under constant acceleration

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SUMMARY

The discussion focuses on calculating distance and velocity under constant acceleration, specifically with an acceleration of 11 m/s². After 6 seconds, the velocity is confirmed to be 66 m/s, and the distance traveled is calculated using the formula d = (1/2)at², resulting in 198 meters. The participants clarify that displacement is equivalent to distance in this context and emphasize the importance of using the correct formulas for such calculations.

PREREQUISITES
  • Understanding of kinematic equations
  • Familiarity with the concept of constant acceleration
  • Knowledge of basic physics terminology
  • Ability to perform algebraic manipulations
NEXT STEPS
  • Study the kinematic equation d = vt + (1/2)at²
  • Learn about the relationship between displacement and distance in physics
  • Explore examples of constant acceleration scenarios
  • Investigate the implications of average velocity in uniformly accelerated motion
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding motion under constant acceleration.

Alpha[X]²
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A body starts from rest with an acceleration of 11m/s².

What is it's velocity after 6 seconds. (I got 66 for that) and how far did it travel?

What formula do I have to use for this?
 
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There is a well-known expression which represents displacement under constant acceleration. Try to find it. (Either here or google it up.)
 
Is displacement the same as distance?
 
Alpha[X]² said:
Is displacement the same as distance?

Yes it is, in this case.
 
What about the first part of the question?

I tried to average out the velocity and then using the d=vt formula I got 198m.
 
Alpha[X]² said:
What about the first part of the question?

I tried to average out the velocity and then using the d=vt formula I got 198m.

Yes, you did it correctly. Remember that you could calculate the displacement averaging out the velocity only because the acceleration is constant.

The other formula would be: d = (1/2)at2. Check that the solution is the same.

Try to show that v(average)*t = (1/2)at2
 
Last edited:

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