How Does Exponential Decay Relate to Velocity in Differential Equations?

[konichiwa2x]

The motion of a body is given by the equation dV(t)/dt = 0.6 - 3V(t)
where V(t) is the speed (in m/s) at time t (in second). If the body was at rest at t = 0

1) What is the magnitude of the inital acceleration?
2) The speed of the body varies with time as

(A) [tex](1 - e^-^3^t) [/tex]
(B) [tex]2(1 - e^-^3^t)[/tex]
(C) [tex]\frac{2}{3}(1 - e^\frac{-3t}{2})[/tex]
(D) [tex]\frac{2}{3}(1 - e^\frac{-3t}{3})[/tex]

(B) is the correct answer for Q(2) . But how do you arrive at it? And how did they manage to get a 'e' in the answer?

Please help.

 

[Päällikkö]

Please show what you've tried.

You're dealing with a separable differential equation, do you know how to solve one?

 

[konichiwa2x]

what is a separable equation? I know basic calculus. but i have no clue on how to arrive at the answer to this question,

 

[Päällikkö]

I googled for "separable differential equation" and found a decent looking text:
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/moreApps/separable.html

Applying the above to your problem:

[tex]\frac{dV(t)}{dt} = 0.6 - 3V(t)[/tex]

[tex]dV(t) = (0.6 - 3V(t))dt[/tex]

[tex]\frac{dV(t)}{0.6 - 3V(t)} = dt[/tex]

[tex]\int_{V_0}^{V}\frac{dV(t)}{0.6 - 3V(t)} = \int_{t_0}^t dt[/tex]

(You could also use indefinite integral, and solve for the C with the information given in the problem ie. "body was at rest at t = 0")Can you manage the rest?PS. There's something wrong with the equation or the correct answer. With the given equation you should arrive at:
[tex]0.2(1-e^{-3t})[/tex]

To get the given answer (B), the original equation should be:
[tex]\frac{dV(t)}{dt} = 6 - 3V(t)[/tex]

 

[HallsofIvy]

If you do not know how to solve differential equations, and presumbably aren't expected to here, sSince you are given 4 possible functions, work the other way. Plug each into the equation of motion and see which works. ([itex]\frac{dV}{dt}= 0.6- 3V[/itex] won't work with any of them- as said, it must be 6- 3V.)

As for part A, that's easy. Just evaluate [itex]\frac{dV}{dt}= 0.6- 3V[/itex] at t= 0. (Of course, you are told V(0).)

 

[konichiwa2x]

[tex]\int_{V_0}^{V}\frac{dV(t)}{0.6 - 3V(t)} = \int_{t_0}^t dt[/tex]

(You could also use indefinite integral, and solve for the C with the information given in the problem ie. "body was at rest at t = 0")

I am not very sure on how to procede from here.(I have just started learning calculus last week). Anyway should I use
[tex] \int uv =u \int v - \int \frac{(du)}{(dx)}\int v [/tex] rule?

 

[Päällikkö]

I am not very sure on how to procede from here.(I have just started learning calculus last week).

Here's a formula that should help you:

[tex]\int \frac{dx}{x} = \ln |x| + C[/tex]