Linear acceleration as a fx of displacement

[Nirck]

Hi everyone. I just began my Calculus based physics course 3 weeks ago and I really love it. I've been going ahead of the class though and came across the following problem:

As g, the acceleration due to gravity, is constant for only a limited range of height distances, a better approximation would be the linear equation

a(h) = g - hg'

where h is the distance from the ground and g' is a small constant of proportional dimensions.

My problem is that I now need to derive velocity as a function of distance, and then as a function of time.

I've tried everything I can think of, but I just can't seem to wrap my head around how I can express height as a function of time, or vice versa, without knowing either acceleration as a function of time or velocity as a function of time, but I can't get either of those without knowing one of the other two!

Any help would be greatly appreciated. Thanks in advance.

 

[HallsofIvy]

Hi everyone. I just began my Calculus based physics course 3 weeks ago and I really love it. I've been going ahead of the class though and came across the following problem:

As g, the acceleration due to gravity, is constant for only a limited range of height distances, a better approximation would be the linear equation

a(h) = g - hg'

where h is the distance from the ground and g' is a small constant of proportional dimensions.

Actually, a(h)= GM/(h+ R)² where M is the mass of the earth, R is the radius of the Earth and G is the "universal gravitational constant" but yes, that is a better approximation for a(h) than just a constant.

My problem is that I now need to derive velocity as a function of distance, and then as a function of time.

I've tried everything I can think of, but I just can't seem to wrap my head around how I can express height as a function of time, or vice versa, without knowing either acceleration as a function of time or velocity as a function of time, but I can't get either of those without knowing one of the other two!

Any help would be greatly appreciated. Thanks in advance.

With continuously varying quantities, acceleration is the derivative of v, with respect to time: a= dv/dt. By the chain rule, then a= (dv/dx)(dx/dt)= v(dv/dx). To find the acceleration as a function of height, x, you would need to solve the differential equation v(dv/dx)= g- g' x. That can be "separated' as dv= (g- g'x)dx and integrated directly.

 

[Nirck]

To find the acceleration as a function of height, x, you would need to solve the differential equation v(dv/dx)= g- g' x. That can be "separated' as dv= (g- g'x)dx and integrated directly.

I was able to get to this point myself, but I'm confused as to where the v goes in this equation. It seems to me that the equation would be v(dv) = (g0 - g'x)(dx), and I don't quite understand how the v(dv) would represent velocity as a function of height...??

 

[HallsofIvy]

I was able to get to this point myself, but I'm confused as to where the v goes in this equation. It seems to me that the equation would be v(dv) = (g0 - g'x)(dx), and I don't quite understand how the v(dv) would represent velocity as a function of height...??

You integrate, as I said. $$\int vdv= \int (g_0- g' x)dx$$. You will get
[tex]\frac{1}{2}v^2= g_0x- \frac{g'}{2}x^2+ C[/itex]. The "C" will depend upon the initial height and velocity.