dynamics
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Post Fiend
Joined: Jun 2004
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From: Joppa, Maryland
dynamics
I can't figure out my hw.
This is for my dynamics class .. the teacher is foreign and difficult to understand.
The acceration of point A is defined by the relation a=600x*(1+kx^2), where a and x are expressed in ft/s^2 and feet, repectively and k is a constant. Knowing that the velocity of A is 7.5 ft/s when x=0 and 15ft/s when x=.45ft, determine the value of k.
thanks for any help. i'll be reading over and over trying to figure it out.
This is for my dynamics class .. the teacher is foreign and difficult to understand.
The acceration of point A is defined by the relation a=600x*(1+kx^2), where a and x are expressed in ft/s^2 and feet, repectively and k is a constant. Knowing that the velocity of A is 7.5 ft/s when x=0 and 15ft/s when x=.45ft, determine the value of k.
thanks for any help. i'll be reading over and over trying to figure it out.
Post Fiend
Joined: Mar 2004
Posts: 24,418
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From: San Francisco, Ca.
That's so easy.
To determine the value of K, you just...........................................
Never mind.
It's better if you figure it out yourself.
How else are you going to learn.....................................Greek.
To determine the value of K, you just...........................................
Never mind.
It's better if you figure it out yourself.
How else are you going to learn.....................................Greek.
Postmaster
Joined: Aug 2004
Posts: 4,004
Likes: 1
From: North of Boston MA
Digging deep into the memory bank here but since you have a relationship between velocity at t(0) and t(1) and you have an equation for acceleration, then you should be able to plug them into the equation
x=V(0)t + .5at^2 and solve for "k" by inserting your equation for "a"....
Let me play some more.....
x=V(0)t + .5at^2 and solve for "k" by inserting your equation for "a"....
Let me play some more.....
Posting Guru
Joined: Jun 2005
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From: Location, Location.
You have :
1) a(x) = acceleration as a function of position x
2) two values for v(x) = velocity as a function of position x
Just as a = dv/dt, and conversely, v(t) = integral(a)dt
When you have a(x),
v(x) = integral(a)dx
= integral(600x + 600kx^2)dx
= 300x^2 + 200kx^3 + d
You know v(0) = 7.5 = d
And v(0.45) = 15 = 300(.45^2) + 200k(.45^3) + 7.5
= 60.75 + 18.225k + 7.5
k = -2.922
No?
1) a(x) = acceleration as a function of position x
2) two values for v(x) = velocity as a function of position x
Just as a = dv/dt, and conversely, v(t) = integral(a)dt
When you have a(x),
v(x) = integral(a)dx
= integral(600x + 600kx^2)dx
= 300x^2 + 200kx^3 + d
You know v(0) = 7.5 = d
And v(0.45) = 15 = 300(.45^2) + 200k(.45^3) + 7.5
= 60.75 + 18.225k + 7.5
k = -2.922
No?
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Postmaster
Joined: Aug 2004
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From: North of Boston MA
Originally Posted by fred_79f250
When you have a(x),
v(x) = integral(a)dx
= integral(600x + 600kx^2)dx
= 300x^2 + 200kx^3 + d
v(x) = integral(a)dx
= integral(600x + 600kx^2)dx
= 300x^2 + 200kx^3 + d
Using your way with the correct equation and
k = -8.something
Posting Guru
Joined: Jun 2005
Posts: 1,254
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From: Location, Location.
Originally Posted by Nitramjr
.... you should be integrating 600x + 600kx^3 (not squared). Look again at the original equation. I questioned that too.
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