Find the Velocity when accelration is not constant

[Northbysouth]

Find the Velocity when acceleration is not constant

Homework Statement

A particle moves along the positive x-axis with an acceleration in meters per second squared which increases linearly with x expressed in millimeters, as shown on the graph for an interval of its motion. If the velocity of the particle at x = 40 mm is 0.36 m/s, determine the velocity at x = 130 mm.

I have attached an image of the question.

Homework Equations

The Attempt at a Solution

I had thought to use:
v_x= v_0x+at
or
s = s₀ + v_0xt +1/2*at²

But due to the acceleration not being constant these equations are not suitable. I'm guessing that I need an integral but I'm not sure how to start. Any help would be appreciated.

 

[Doc Al]

Consider the work done by the force that is accelerating the particle.

 

[rude man]

Well, you have a(x) = bx with b given.
You also have a = dv/dt = dv/dx * dx/dt = v dv/dx
So v dv/dx = bx.

This diff. eq. can easily be solved by separation of variables. Remember the constant of integration which you determine from the given initial condition ...

 

[voko]

Well, you have a(x) = bx with b given.

You surely meant a(x) = bx + c.

 

[rude man]

You surely meant a(x) = bx + c.

I surely did.

Same approach, though.

 

[Northbysouth]

I've only recently started learning about differential equations so bear with me.

From what I understand I should get:

v dv/dx = bx + C

v dv = (bx + C)dx

∫v dv = ∫(bx + C)dx

v²/2 = bx²/2 + Cx + D

I then know that v(0.040 meters) = 0.36 m/s and that I can use this information to find D but how do I find b and D? Is b just the gradient of the slope?

 

[voko]

Pick any two values of x. Then, since a = bx + c, you should get two linear equations for b and c with known values of a and x. Solve them.

 

[rude man]

I've only recently started learning about differential equations so bear with me.

From what I understand I should get:

v dv/dx = bx + C

v dv = (bx + C)dx

∫v dv = ∫(bx + C)dx

v²/2 = bx²/2 + Cx + D

I then know that v(0.040 meters) = 0.36 m/s and that I can use this information to find D but how do I find b and D? Is b just the gradient of the slope?

Yes, voko has shown you how you can get b and c. Remember y = (slope)*x + y-intercept from high school algebra? In this case b is the slope and c is the a-intercept.

You did well with solving the diff. eq. BTW.

 

[Northbysouth]

I got it. The answer is 1.0366 m/s.

Thank you to everyone for your help.

 

[Doc Al]

Just for fun, here's how I would have solved it using the method I hinted at in my earlier post:

ΔKE = ∫F.dx
Δ(v²/2) = ∫a.dx

The integral is the area under the curve, so you can just read it off of your diagram. No need for any formal calculus.

The methods are entirely equivalent, of course. And until you are comfortable with the math, probably best to work things out as you did.