Why is the integral of a(x) different from the integral of a(t)?

[Elias Waranoi]

Hello! First time poster, please treat me well! I've already solved the problem below on my second attempt with the help of kinetic energy but I want to know why my first attempt gives a wrong answer.

1. Homework Statement
A force in the +x-direction with magnitude F(x) = 18.0 N - (0.530 N/m)x is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?

Homework Equations

a = F/m

The Attempt at a Solution

F(x) = 18.0 - 0.530x =>
a(x) = 18.0/m - 0.530x/m =>
Integrate =>
v(x) = 18.0x/m - 0.530x²/2m = 3x - 0.530/12 x²
v(14) = 33.3

Why is this incorrect? I'm guessing that the integral of a(t) is velocity but the integral of a(x) is something else.

 

[berkeman]

Welcome to the PF

a(x) = 18.0/m - 0.530x/m =>
Integrate =>
v(x) = 18.0x/m - 0.530x2/m

When you integrate x, you don't just get x^2...

 

[Elias Waranoi]

Welcome to the PF When you integrate x, you don't just get x^2...

Oops, I edited my question. The result is still wrong though.

 

[berkeman]

v(x) = 18.0x/m - 0.530x2/2m = 3x - 0.530/12 x2
v(14) = 33.3

Looks like it's just a math error when you plug in x=14. Try again?

 

[Elias Waranoi]

Looks like it's just a math error when you plug in x=14. Try again?

I typed this into wolfram alpha "3*14 - 0.53/12*14^2" and got 33.3. When I use W = K₂ - K₁, W = F*s with the average force from x = 0 to x = 14 I get the velocity at 14 meters to be 8.17 which is the correct answer.

 

[Doc Al]

I'm guessing that the integral of a(t) is velocity but the integral of a(x) is something else.

Exactly. If you integrate F(x)dx, you can calculate the work and thus the change in KE.

 

[berkeman]

"3*14 - 0.53/12*14^2"

You should use parenthesis, I believe. x^2 is not in the denominator...

 

[berkeman]

Exactly. If you integrate F(x)dx, you can calculate the work and thus the change in KE.

Ah, so the choice of integral is wrong, regardless of errors in the integration. Thanks Doc.

 

[Elias Waranoi]

Exactly. If you integrate F(x)dx, you can calculate the work and thus the change in KE.

Ahh, so if F(x)dx integrated is work I'm guessing that integrating a unit1 with unit2 equals unit1*unit2. So when I integrated a(x)dx I actually got meter²/time² instead of velocity. Is this correct?

 

[Doc Al]

Ahh, so if F(x)dx integrated is work I'm guessing that integrating a unit1 with unit2 equals unit1*unit2. So when I integrated a(x)dx I actually got meter²/time² instead of velocity. Is this correct?

That's true.

 

[haruspex]

the integral of a(t) is velocity but the integral of a(x) is something else.

It is easy to get confused here.
It depends whether you are thinking of a as the physical entity acceleration or as purely a mathematical function.
As a physical entity, it's not whether you think of a as a function of time or of displacement that matters. What matters is what you integrate with respect to, as you figured out. Integration wrt a variable is akin to multiplication by that variable. ∫a.dt gives Δv, while ∫a.dx gives Δ(v²)/2.
If you are given a as a function of displacement, ∫a.dt still yields velocity, but how are you to perform the integral? E.g. if told a=kx³, your integral is ∫kx³.dt. Since you do not know what x is as a function of t this is not directly solvable. If you are told the total displacement then you can use ∫a.dx instead to find the velocity, but what if you are only told the elapsed time? In that case you would have to write out and solve the differential equation.