Help - finding what slope means (1 Viewer)

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urgent help - finding what slope means

Suppose a student is given position-time data for a cart which was known to have an initial velocity of zero. if the student plots the fifth root of time (t^1/5) on the y axis, what variable should he plot on the x axis so that the resulting graph is a straight line? how is the slope of the line related to the acceleration of the cart?


[tex]d = \frac{1}{2}at^2[/tex]
[tex]d^\frac{1}{10} = (\frac{1}{2}a)^\frac{1}{10} t^\frac{1}{5}[/tex]

this is what i've done so far.... can anyone give me some hints as to what to do next? thank you i need to know the answer in 1 hour =(
 

Astronuc

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Normally if one was given position - time data, one would plot position (displacement from some reference) on the y axis (a dependent variable) and time on the x-axis (independent variable).

The slope would then be the speed or magnitude of velocity, and it would be a straight line if there were no acceleration.

If there were an acceleration or deceleration the slope would not be straight. The second derivative of the curve would give the acceleration. If acceleration were constant, distance would be a parabolic function of time.

To get a straight line in a plot, one would have to take [tex]d = \frac{1}{2}at^2[/tex] and take the log or ln of the expression to get
[tex]ln d = ln\frac{a}{2} + 2 ln t[/tex] and this assumes a is constant.

Think about y = t1/5

x = ?, and one wants to find dy/dx = d(t1/5)/d(?)

to get a straight line [itex] \frac{d^2y}{dx^2}[/itex] = 0
 

Gokul43201

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an_mui said:
Suppose a student is given position-time data for a cart which was known to have an initial velocity of zero. if the student plots the fifth root of time (t^1/5) on the y axis, what variable should he plot on the x axis so that the resulting graph is a straight line? how is the slope of the line related to the acceleration of the cart?
[tex]d = \frac{1}{2}at^2[/tex]
[tex]d^\frac{1}{10} = (\frac{1}{2}a)^\frac{1}{10} t^\frac{1}{5}[/tex]
this is what i've done so far.... can anyone give me some hints as to what to do next? thank you i need to know the answer in 1 hour =(
You've got the answer right there : call [itex]t^{1/5} = y [/itex]. Then rearranging the equation gives

[tex]y = d^{1/10}/(a/2)^{1/10} = k \cdot d^{1/10} [/tex]

So, if you make the x-axis contain values of [itex]d^{1/10} [/itex] you are done. You will then have a line of the form [itex]y = kx [/itex], where the slope of the line, k, is ... ?
 
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