Finding Position with a Quadratic Velocity

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SUMMARY

The discussion focuses on calculating the position and velocity of a particle under a time-dependent acceleration defined as ax = (0.1 m/s³)t. The initial conditions specify a velocity of 10.0 m/s to the right and a position of -4.8 m. The correct velocity function is established as v(t) = (0.05t²) + 10. The position function x(t) = (0.017t³) + 10t + 4.8 was initially deemed incorrect, prompting inquiries about potential errors related to significant figures.

PREREQUISITES
  • Understanding of kinematics and motion equations
  • Familiarity with calculus concepts, specifically integration
  • Knowledge of significant figures in scientific calculations
  • Ability to analyze particle motion in one dimension
NEXT STEPS
  • Study the principles of integration in physics to derive velocity and position functions
  • Review significant figures and their importance in physics calculations
  • Explore the concept of average velocity and its calculation methods
  • Investigate the relationship between instantaneous and average velocities in particle motion
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Students studying physics, particularly those focusing on kinematics and motion analysis, as well as educators seeking to clarify concepts related to particle dynamics and significant figures.

glockstock
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Homework Statement


In the time interval from 0.0 s to 10.8 s, the acceleration of a particle traveling in a straight line is given by ax = (0.1 m/s3)t. Let to the right be the +x direction. The particle initially has a velocity to the right of 10.0 m/s and is located 4.8 m to the left of the origin.


Homework Equations


(a) Determine the velocity as a function of time during the interval.

(b) Determine the position as a function of time during the interval.

(c) Determine the average velocity between t = 0.0 s and 10.8 s.
3 m/s

Compare it to the average of the instantaneous velocities at the start and ending times. Are these two averages equal?



The Attempt at a Solution



(a) v(t) = (0.05t^2)+10 this was determined to be right...

(b) x(t) = (0.017t^3)+10t+4.8 this is wrong...why? Significant Figures?
 
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glockstock said:

Homework Statement


In the time interval from 0.0 s to 10.8 s, the acceleration of a particle traveling in a straight line is given by ax = (0.1 m/s3)t. Let to the right be the +x direction. The particle initially has a velocity to the right of 10.0 m/s and is located 4.8 m to the left of the origin.


(b) x(t) = (0.017t^3)+10t+4.8 this is wrong...why? Significant Figures?

that means for t=0, x= -4.8m (+ve is to the right means -ve is to the left)
 

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