How to convert a slope to an acceleration?

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SUMMARY

The discussion focuses on converting the slope of a graph representing position versus time squared (x(t²)) into acceleration. The participant calculated the slope as approximately 4.708 using the formula Δy/Δx, where Δy is the final position (0.610m) and Δx is the final time squared (0.1296s²). The key takeaway is that the slope of the x(t²) graph represents half of the acceleration (a), as derived from the equation x = (1/2) a t². Therefore, the acceleration can be determined by multiplying the slope by 2, yielding an acceleration of approximately 9.416 m/s², consistent with the expected gravitational acceleration of 9.81 m/s².

PREREQUISITES
  • Understanding of basic physics concepts, specifically kinematics.
  • Familiarity with graphing and interpreting position versus time graphs.
  • Knowledge of the equation of motion for constant acceleration: x = (1/2) a t².
  • Basic mathematical skills for calculating slopes (Δy/Δx).
NEXT STEPS
  • Study the relationship between position, velocity, and acceleration in kinematics.
  • Learn how to derive equations of motion from graphical representations.
  • Explore the concept of constant acceleration and its implications in real-world scenarios.
  • Practice plotting and interpreting various kinematic graphs, including x(t) and x(t²).
USEFUL FOR

High school physics students, educators teaching kinematics, and anyone interested in understanding the relationship between position, time, and acceleration in motion.

Ellio
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Homework Statement
"From the slope (of the graphic) you calculated, determine the acceleration of the object."
Relevant Equations
Slope: Δy/Δx
Gravitational acceleration on Earth ≅ 9.81 m/s²
Hello, I hope you are all very well.
I am in second year of High School and I have a practical work in physics.
The experiment was to release a long tape with a mass of 40g at the end from a certain height. An instrument would hit the tape 50 times/s and put a mark each time. From that we have to complete a series of exercises.

One of the exercise is however causing me some trouble. We had to write down the different position of all the mark that are on the tape on a graphic. The first one is a x(t) graphic and the second one x(t²).
I got a curved line with the first one:
1571916916886703887411023115953.jpg

and a straight line with the x(t²) graphic:
20191024_133823.jpg


Here's my problem... We're told to first calculate the slope of this (x(t²))graphic. The expression to calculate a slope being Δy/Δx. The very last "position" (y axis) on the graphic is 0.610m at 0.1296s² (x axis).
0.610/0.1296 ≃ 4.708
Assuming that the slope = 4.708, how could I know find the acceleration from that ? The graphic being a constant straight line, isn't the acceleration = 0 ?
But that's not the case (a=9.81), if I'm not mistaken...

I am really sorry to bother you with a such a stupid question ! I really thank you in advance.

(ps.: I'm Swiss, sorry for my English...)
 
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In the first graph, you plot x vs t, and you see the slope, which is Δx/Δt increasing. In the second graph, you plot x vs t^2, so the slope is Δx/Δ(t^2). Since an object moving with constant acceleration a follows the equation x = \frac{1}{2} a t^2, the slope of this line (Δx/Δ(t^2))is 1/2 a. This is just what you found. Does this help?
 
phyzguy said:
In the first graph, you plot x vs t, and you see the slope, which is Δx/Δt increasing. In the second graph, you plot x vs t^2, so the slope is Δx/Δ(t^2). Since an object moving with constant acceleration a follows the equation x = \frac{1}{2} a t^2, the slope of this line (Δx/Δ(t^2))is 1/2 a. This is just what you found. Does this help?
Absolutely, it really helped me thank you a lot !
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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