Tough Physics integration problem dealing with kinematics.

AI Thread Summary
A train with an initial speed of 22.0 m/s brakes using a velocity-dependent acceleration formula. The challenge is to determine the distance traveled until it stops, requiring integration of the equation (v+u)dv = (α)dt. Participants suggest eliminating time in favor of distance by expressing dt in terms of dx. The discussion emphasizes finding a relationship between velocity and distance rather than time to simplify the integration process. This approach aims to facilitate the calculation of the stopping distance effectively.
lorenzosounds
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Homework Statement



A train traveling at v0 = 22.0 m/s begins to brake by applying a velocity-dependent instantaneous acceleration a(v) = α /(v + u) m/s2, where α = – 23.0 m2/s3, v is the instantaneous velocity of the train, and u = 0.5 m/s. Determine the distance traveled by the train until it comes to a complete stop.

I know I need to integrate, but I don't know how to get to d(t).

Homework Equations



(v+u)dv = (α)dt

The Attempt at a Solution



-22v-11 = 23.0(dt) is where I'm at. I'm not sure how to integrate both sides. How do I get to v(t), so I can solve d(t)?
 
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lorenzosounds said:
2. Homework Equations [/b]
(v+u)dv = (α)dt

Hello, lorenzosounds. You'll need to eliminate time, t, in favor of distance, x. Can you express dt in terms of v and dx?
 
TSny said:
Hello, lorenzosounds. You'll need to eliminate time, t, in favor of distance, x. Can you express dt in terms of v and dx?


Would it be -(1/2v^2+0.5v)=-23.0t ?
 
lorenzosounds said:
Would it be -(1/2v^2+0.5v)=-23.0t ?


No. If you integrate (v+u)dv = (α)dt, then the left side would be integrated from vo to v. So, vo as well as v would appear on the left side.

But, then you'll have to solve this for v as a function of t, which will get messy. After that you would have to integrate one more time to find the distance traveled.

There's an easier way. Instead of trying to get v as a function of t, try to get v as a function of x. Go back to (v+u)dv = (α)dt. Is there a way to express the differential dt in terms of dx? Hint: v = dx/dt.
 
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