Acceleration proportional to velocity

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SUMMARY

The discussion centers on a physics problem involving a particle accelerated from rest with acceleration proportional to its velocity. The key equations established include a = kv, where k is a constant, leading to the integration of dv/v = kdx. After traveling 10 meters at a speed of 35 m/s, the goal is to determine the speed after traveling 20 meters. The participants confirm the validity of the approach and emphasize the importance of understanding the implications of dividing by velocity.

PREREQUISITES
  • Understanding of basic calculus and integration techniques
  • Familiarity with the concept of acceleration and its relationship to velocity
  • Knowledge of differential equations and their applications in physics
  • Basic principles of kinematics in one-dimensional motion
NEXT STEPS
  • Study the derivation of the equation a = kv in the context of exponential growth
  • Learn about integrating differential equations in physics applications
  • Explore the implications of dividing by variables in mathematical equations
  • Investigate real-world applications of acceleration proportional to velocity in mechanics
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Students studying physics, particularly those focusing on kinematics and dynamics, as well as educators looking for examples of acceleration concepts in motion.

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Homework Statement


A particle is accelerated from rest with acc. proportional to velocity. After traveling 10 meters the speed is 35 m/s, find the speed after the particle traveled 20 meters. (Changed the numbers a bit from the original so the answer ppbly will not be clean.)

Homework Equations



a=v dv/dx possibly

The Attempt at a Solution


Sat on this one for a long, long time

Tanks in advance
 
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"Tanks"?

No, a is not "v dv/dx". Rather a= dv/dx which, here, is proportional to v:
a= dv/dx= kv for some number k.

From that dv/v= kdx. Now integrate both sides.
 
a = dv/dt, I think you mean?
 
Yea, watch out for those advancing tanks; hah, no I meant to say thanks.
Anyway, I do believe that a= v dv/dx ; dv/dt = (dv/dx)(dx/dt) = v (dv/dx). Does this seem reasonable:
a=kv=v dv/dx
k=dv/dx
k dx = dv
intg.
v=kx
? or is dividing by v not allowed => loss of roots or something?
 
I do believe you're right, a = v dv/dx. (In one dimension of course) And I don't see anything wrong with your work.

Can you think of a reason why dividing by v would not be allowed? (hint: what number(s) are you not allowed to divide by?)
 

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