How Do You Calculate Position from a Velocity Graph?

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SUMMARY

The discussion focuses on calculating the position of a car based on its velocity graph, starting from an initial position of 10 meters at time zero. At 2 seconds, the car's velocity is 4 m/s, leading to a calculated position of 18 meters using the formula x = vt + x0. The conversation highlights the importance of understanding that the negative slope of the velocity graph indicates deceleration and potential direction changes. Integration is suggested as a method to derive position from the velocity function over time.

PREREQUISITES
  • Understanding of basic kinematics, specifically velocity and position relationships.
  • Familiarity with the concept of integration in calculus.
  • Knowledge of interpreting velocity graphs and their implications on motion.
  • Ability to apply the equation V = d/t in practical scenarios.
NEXT STEPS
  • Study the fundamentals of integration and its application in physics.
  • Learn how to analyze velocity vs. time graphs for motion analysis.
  • Explore the concepts of acceleration and deceleration in kinematics.
  • Review examples of calculating position from varying velocity functions.
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for practical examples of motion analysis using velocity graphs.

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Homework Statement



5. A car starts from xi = 10m at ti = 0s and moves with the velocity graph shown in figure on the right.
a. What is the object’s position at t = 2s, 3s, and 4s?
b. Does this car ever change direction? If so, at what time?

Homework Equations



V = d/t

The Attempt at a Solution



for t at 2 secs...
according to graph v= 4m/s

so position (d) = (t)(v)
= (2)(4)
= 8m?

am i doing it right?


OR

v = change d/change t
4m/s = (x-10)/(2-0)
4m/s = (x-10)/(2)
8m/s = x-10
x = 18m

is that correct?
 

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for t at 2 secs...
according to graph v= 4m/s

so position (d) = (t)(v)
= (2)(4)
= 8m?

am i doing it right?
Well this can't be right if the car starts at 10 m and adds distance. The slope of the plot of velocity vs time is negative, which indicates the car is decelerating, and the since the slope is constant, the deceleration is constant.

Certainly when the car has a negative velocity, it is reversing.

Is one familiar with integration?

if v(t) = d x(t)/dt, then

x(t) = [tex]\int_0^t\,v(t) dt\,+\,x(0)[/tex]

This might be useful:

http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html
 

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