The slope of this line is 36.9. However, because we know the following equation:
My question is, does this mean that the acceleration the the object is (36.9 * 2) = 73.8, or is the acceleration just simply walking to be 36.9?
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edited Feb 26 "16 in ~ 2:37
request Feb 26 "16 in ~ 2:24
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Worked with in comments, however a recap.
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Neither exactly, this graph isn"t very useful for analysis off acceleration, as the slope will not it is in constant. What friend have effectively plotted is $y(x)=v\sqrtx+\fracax2$. Eyeballing this look at linear, yet it really isn"t. Using any information from your fit wouldn"t assist since friend tried to fit a linear relation, rather of this peculiar one. Go ago and refit the points through the equation I provided you, fitting for "a" and "v", then "a" is merely the acceleration. Utilizing the direct slope in this situation is useless, external a stormy guess.
To correctly solve for acceleration, an initial you need to discover out the velocity together a role of time, then you can discover the acceleration. You can skip the if you desire a computer system to right the data for you, through the prescribed equation. However, i still recommend simply fitting the street verse time v a quadratic (if acceleration is uniform), and taking a derivative twice or analysis off the equipment value. If her acceleration isn"t continuous you can use various other fits (cubic, sinusoidal, hyperbolic, etc.), which version what you believe is happening. Be warned the you can"t just fit data with an arbitrary fit and also expect to acquire the truth. Everything fit girlfriend use, is the model you are using, and also the presumptions you are making. Additionally note in ~ 4 data points girlfriend are limited to what fits you deserve to make, e.g., you can"t ask because that a fifth order right with any sincerity.
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Is there any kind of hope for this graph to gain you acceleration? Roughly. You have the right to approximate the acceleration for $t\gg 1$, due to the fact that we deserve to ignore the first term through $x \gg \sqrtx$ (equally if you think apriori the $a \gg v$). Then the equation is roughly, $y(x)\approx \fraca2x$. Therefore $\fraca2\approx$ slope, then $a=2\times$ slope. This makes sense, as what is physically keep going is as time walk on for a lengthy time, the acceleration hatchet is really overcoming the displacement, and that early velocity is tiny compared to exactly how much velocity has actually been included due come the acceleration.
Let me give you an example of the accuracy here. Say the true values room $v_0=5$ and also a=3. Take measurements at $x$= 2, 4, 9,998, and 10,000 (measurements at time equals $\sqrt2$, 2, $\sqrt9,998$ and also 100). The values space y(2)=10.0711, y(4)=16, y(9,998)=15,496.9, and also y(10,000)=15,500. Then the slope between 2 and also 4 is $\sim$2.96 for this reason acceleration of about 5.93. The slope between 10,000 and also 9,998 is 1.525, and also acceleration of approximately 3.05. You deserve to see the the values of 2 and 4 give pretty negative resolution, nearly 100% error, i beg your pardon is close come what your resolution is.