Tough Physics integration problem dealing with kinematics.

Click For Summary

Homework Help Overview

The problem involves a train decelerating with a velocity-dependent acceleration, requiring integration to find the distance traveled until it stops. The subject area is kinematics, focusing on the relationship between velocity, acceleration, and distance.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to integrate the given equations and question how to eliminate time in favor of distance. There are attempts to express the relationship between differentials and velocity.

Discussion Status

Some participants have offered guidance on expressing time in terms of distance, suggesting alternative approaches to the integration process. Multiple interpretations of the integration steps are being explored without a clear consensus on the best method.

Contextual Notes

Participants are navigating the complexity of integrating velocity-dependent acceleration and are considering the implications of expressing variables in different forms. There is an acknowledgment of the potential messiness involved in solving for velocity as a function of time.

lorenzosounds
Messages
4
Reaction score
0

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)?
 
Physics news on Phys.org
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.
 

Similar threads

Replies
3
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K