A boat's acceleration is proportional to its velocity

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SUMMARY

The discussion focuses on the physics of a 1000kg boat decelerating from 90 km/h to 45 km/h after its engine is shut off. The frictional force acting on the boat is modeled as f = 70v, where v is the velocity in meters per second. The problem involves solving a differential equation of the form dv/dt = -kv, which describes the relationship between acceleration and velocity. The conversation highlights the use of calculus techniques, specifically separation of variables, to find the velocity function over time.

PREREQUISITES
  • Understanding of Newton's second law (f = ma)
  • Basic knowledge of differential equations
  • Familiarity with calculus concepts, particularly integration
  • Concept of drag force in fluid dynamics
NEXT STEPS
  • Study the method of separation of variables in differential equations
  • Learn about the derivation and application of drag force equations, specifically for water and air
  • Explore the implications of exponential decay in velocity over time
  • Investigate the effects of varying coefficients in drag force equations, such as Cd and A
USEFUL FOR

Physics students, engineering students, and anyone interested in the dynamics of motion and fluid resistance in real-world applications.

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A 1000kg boat is traveling at 90km/h when its engine is shut off. The magnitude of the frictional force f between the boat and water is proportional to the speed v of the boat: f=70v where v is in meters per second and f is in Newtons. Find the time required for the boat to slow to 45 km/h.

This problem wasn't assigned so I might be trying something I'm not supposed to know how to do. I have a FBD and have defined the x-axis in the direction of the boat's motion.

-f=ma=-70v
therefore v(t)=-\int .070vdt

I wish I could show more work but I'm don't know where to go next. I think this employs some calculus I'm not familiar with so if someone could just point out a concept I need to look at I would appreciate it.
 
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Write a=(dv/dt).
So, you have (dv/dt)=-kv, where k is a constant.

This is a "differential equation for v".
To solve for v, you have to find a function v that satisfies this equation. You don't need a class in differential equations to solve this, however.

Can you think of a function of t whose derivative is proportional [with a negative constant] to itself? If you can't you can try a technique called "separation of variables" to obtain such a function.
 
Thanks I got it.
 
To keep the problem real use drag force proportional to the velocity squared as:

Water Drag = 1/2 xCd x A x V^2 or proportional to V^2 with Cd and A constant.

Same functional relation for air drag except different Cd and A
 
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