How to Find g from the Gradient of a Distance-Time Graph?

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SUMMARY

The discussion focuses on calculating the gravitational acceleration (g) from the gradient of a distance-time graph for an object dropped from rest. The user derived the formula g = 2h / t², where h is the height and t is the time squared. They calculated the gradient of the graph as 4.9 using the formula (Difference y) / (Difference x) and concluded that g can be expressed as g = Gradient x 2, leading to a value of 9.8 m/s². This demonstrates the direct relationship between the gradient of the graph and the acceleration due to gravity.

PREREQUISITES
  • Understanding of kinematic equations, specifically h = g x t² / 2
  • Knowledge of how to calculate the gradient of a graph
  • Familiarity with distance-time graphs
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study the derivation of kinematic equations in physics
  • Learn how to analyze distance-time graphs in detail
  • Explore the concept of acceleration and its calculation from graphs
  • Investigate real-world applications of gravitational acceleration
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Students studying physics, particularly those focusing on kinematics and the principles of motion, as well as educators looking for examples of graph analysis in teaching gravitational concepts.

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Homework Statement


I've been given the task of analysing data from an experiment where an object was dropped with an initial velocity of 0.I've calculated the time in s2 from the original milisec.

Distance Time s2
0.1 0
0.2 0.040804
0.4 0.081796
0.6 0.1225
0.8 0.163216
1 0.204304
1.2 0.245025
1.4 0.286225
1.6 0.326041
1.8 0.367236
2 0.408321

Homework Equations



I was h =g x t2 /2


Which I rearranged to make g the subject g = 2h / 2

The Attempt at a Solution



I've calculated the gradient of the graph as 4.9 using (Difference y) / (Difference x)



Picking 2 points from the graph

I'm stuck at the point where it says "Show how the gradient is used to find a value for g"

Using the equation I've calculated g to be 9.8 but I'm not sure how this related to the gradient apart from being double the amount?

Thanks
 
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What do you mean? You said that you calculated the gradient. Why that isn't a "use of gradient"?
 
Well if I'm using the formula for g(acceleration value)

g = 2h / t2

If the gradient is found by h / t2

Does that mean I can say that g = Gradient x 2

Do you think this answers the question "Show how the gradient is used to find a value for g?"

Thanks
 
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