Acceleration Dependent on Velocity

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SUMMARY

The discussion focuses on the dynamics of a car subjected to a constant force F and an opposing force proportional to its velocity, represented as -kv. The horizontal acceleration is derived as a = (F - kv) / m, where m is the mass of the car. The user attempts to integrate this equation to express the velocity as a function of time but realizes that the integration process requires careful handling of the variable terms. The correct approach involves recognizing that the integration of velocity with respect to time does not yield a simple multiplication of velocity and time.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Basic knowledge of calculus, specifically integration
  • Familiarity with the concept of variable forces in physics
  • Concept of differential equations in motion analysis
NEXT STEPS
  • Study the integration of variable forces in classical mechanics
  • Learn about differential equations and their applications in motion
  • Explore the concept of damping forces and their effects on motion
  • Investigate the relationship between acceleration, velocity, and time in non-constant acceleration scenarios
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the dynamics of motion under variable forces.

mickeylutz
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A constant Force F is applied by the engine of a stationary car causing it to move to the right. Opposing force F, is a force of magnitude -kv, where k is a constant and v is the velocity of the car. The car has a mass of m.

a) Determine the horizontal acceleration in terms of k, v, F and m

b)Derive the equation expressing the velocity of the car as a funtion of time t in terms of k, F, and m.I tried using F=ma:

F-kv=ma
a= (F-kv)/m

Then I integrated this from 0 to t and got

v=(F-kv)t/m

I know that these answers can't be right, since this would be the same result if I had used the equation for constant acceleration. And its not constant acceleration, I'm not sure how to do this.

Thanks!
 
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When you integrate the velocity with respect to time, the result isn't velocity multiplied by time.
 

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