Solve Kinematics Problem Homework: Part B

AI Thread Summary
The discussion centers on solving a kinematics problem involving a differential equation for velocity, v, where acceleration is expressed as a function of velocity, specifically as a0 + kv. The initial attempt at integration leads to confusion about the assumptions made regarding acceleration being constant. Participants suggest that kv should not be simplified to a constant "a" since it varies with velocity. A substitution method is recommended to simplify the equation, leading to a solution involving exponential functions. The conversation emphasizes the importance of correctly handling variable acceleration in the integration process.
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Homework Statement
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The attempt at a solution

I have already solved for part a. It's part b that I'm having trouble with right now.

Here is what I have for part b:

\frac{\mathrm{d} v}{\mathrm{d} t} = a_{0} + kv

dv = a_{0}dt + kvdt

dv = a_{0}dt + adt

\int dv = \int a_{0}dt + adt

And my final answer is:

v = a_{0}t + at

Is this correct?
 
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I don't understand the 3rd step where you replace kv with a.
Are you assuming that the acceleration is kv ?

Also in the final step, you assume that "a" is a constant.
Acceleration can't be constant; it is proportional to the velocity and the velocity is decreasing.

I have forgotten how to solve differential equations, but you might try a substitution to make it look easier. If you let v = u + b where b is a constant, then your dv/dt = ao + kv will change to
du/dt = ao +kb + ku
and you can choose b = -ao/k so
du/dt = ku.
This is the equation for radioactive decay, and it can easily be integrated to get u as an exponential function. Change variables back to v to get an exponential function for v.
 
I made kv = a because the dimension of k is 1/T, where T is time, and v is simply velocity. Therefore, the product of these two variables will be acceleration.
 
But kv is not a constant, so your "a" is not a constant and its integral over time is not at.
 
Does this mean that the integral of adt is equal to v.
 
Your "a" is really "k*v" so its integral over time would be k*distance + constant, where distance is a function of time. I don't think this does any good. Better to leave the kv in the differential equation and make the substitution I suggested in the 2nd post.
 
I don't think your integration is correct. Go to Wolframalpha and use Mathematica's DSolve command for v'[t]=a+k*v[t] I think the solution is v[t]=-a/k + e^(k*t) C1 but not 100% sure since I'm not a regular user of Mathematica
 
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