Determine acceleration from velocity-position plot

[bkw2694]

Homework Statement

The velocity of a particle along s-axis is given as v = 5s^(3/2) , where "s" is in millimeters and v is in millimeters/second. Determine the acceleration when "s" is 2 mm.

So I've been having trouble with differential equations, and I think that's where I'm messing up. My classes haven't really covered velocity-position plots, just velocity-time plots and my book is confusing. The answer to the question is 150 mm/s.

Homework Equations

∫vdv = ∫a ds

The Attempt at a Solution

First I used the equation to try to solve it.
∫(0 to v) vdv = ∫(0 to 2) a ds
This gave me (1/2)v^2 = 2a
Then I plugged v = 5s^(3/2) into it and got (1/2)(25s^3) = 2a
Then I plugged in s = 2 and got 100 = 2a
Then I finished solving for a and got a = 50 mm/s, which is incorrect.

 

[ehild]

Homework Statement

The velocity of a particle along s-axis is given as v = 5s^(3/2) , where "s" is in millimeters and v is in millimeters/second. Determine the acceleration when "s" is 2 mm.

So I've been having trouble with differential equations, and I think that's where I'm messing up. My classes haven't really covered velocity-position plots, just velocity-time plots and my book is confusing. The answer to the question is 150 mm/s.

Homework Equations

∫vdv = ∫a ds

The Attempt at a Solution

First I used the equation to try to solve it.
∫(0 to v) vdv = ∫(0 to 2) a ds
This gave me (1/2)v^2 = 2a

The acceleration is not constant, so its integral with respect to s is not a*s.
It is true (applying chain rule) that a=dv/dt = (dv/ds)(ds/dt)=v dv/ds. Find dv/ds and plug in the formula for v.

 

[bkw2694]

The acceleration is not constant, so its integral with respect to s is not a*s.
It is true (applying chain rule) that a=dv/dt = (dv/ds)(ds/dt)=v dv/ds. Find dv/ds and plug in the formula for v.

Sorry if I'm not following correctly, but are you saying no integral is required for this question?

Solving dv/ds gives me (15/2)s^(1/2). Then multiplying that by the original "v" gives me 5*15/2 (s^2) which after plugging in s = 2 gives me 150mm, the correct answer. Is that the correct setup to this question? And are you saying that I can't use the integrals if acceleration isn't constant?

Thanks in advance!

 

[ehild]

Can you integrate the acceleration if you do not know what function of s it is ?

 

[bkw2694]

Can you integrate the acceleration if you do not know what function of s it is ?

So for me to integrate the acceleration, I need to know a(s)?

 

[insightful]

The answer to the question is 150 mm/s.

gives me 150mm, the correct answer

You are being sloppy with your units. Now is the time to break this habit.

 

[ehild]

So for me to integrate the acceleration, I need to know a(s)?

The integral of a function ∫f(x)dx is a new function, the derivative of which is equal to f(x). F(x)= ∫f(x)dx is called the antiderivative of f(x). The integral of a constant c is cx, but otherwise ∫f(x)dx is not equal to f*x.