Change in velocity when given a graph that shows time and acceleration

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SUMMARY

The discussion focuses on calculating the change in velocity of a stone rolling down a hill, using a graph that depicts acceleration as a function of time. The key equation utilized is v = v0t + 1/2at^2, but the primary method for finding the change in velocity is through the area under the acceleration-time graph. Specifically, the area forms a trapezium between the time intervals of 2.5 seconds and 7.5 seconds, which can be calculated using the trapezium area formula derived from the integral of the acceleration function.

PREREQUISITES
  • Understanding of kinematics, specifically the relationship between acceleration, velocity, and time.
  • Familiarity with graph interpretation, particularly acceleration-time graphs.
  • Knowledge of calculus concepts, specifically integration for area calculation.
  • Ability to apply the trapezium area formula in physics problems.
NEXT STEPS
  • Study the concept of area under a curve in physics, focusing on acceleration-time graphs.
  • Learn how to derive velocity from acceleration using integration techniques.
  • Practice solving kinematics problems involving variable acceleration.
  • Explore the trapezium area formula and its applications in physics calculations.
USEFUL FOR

Students studying physics, particularly those tackling kinematics and integration in their coursework, as well as educators looking for effective methods to explain these concepts.

flynnk567
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Homework Statement



The graph in the figure describes the acceleration as a function of time for a stone rolling down a hill starting from rest.
Find the change in the stone's velocity between 2.5s and 7.5s.
YF-02-43.jpg

Homework Equations



I think v = v0t +1/2at^2

The Attempt at a Solution



I honestly have no idea about how to solve this problem. My professor gives us mastering physics homework problems before he teaches us how to do the material and I'm completely lost right now :/ Any help would be appreciated.
 
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flynnk567 said:

Homework Statement



The graph in the figure describes the acceleration as a function of time for a stone rolling down a hill starting from rest.
Find the change in the stone's velocity between 2.5s and 7.5s.
YF-02-43.jpg

Homework Equations



I think v = v0t +1/2at^2

The Attempt at a Solution



I honestly have no idea about how to solve this problem. My professor gives us mastering physics homework problems before he teaches us how to do the material and I'm completely lost right now :/ Any help would be appreciated.

The change in velocity is the area under the a-t graph. You need to draw the two vertical ordinates in - at 2.5 and 7.5 - to enclose a trapezium; and calculate that.
 
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If you draw out the graph you might notice that your equation a_x = 1*t+2. Drawing lines at x=4.5 and x=7.5 gives you a_x values of 6.5 cm/s^2 and 9.5 cm/s^2. The area of a trapezium can then be calculated by the integral of (1t+2)dt from the lower bound 6.5 to the upper bound 9.5.
 

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