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) (1 - e^-^3^t)
(B) 2(1 - e^-^3^t)
(C) \frac{2}{3}(1 - e^\frac{-3t}{2})
(D) \frac{2}{3}(1 - e^\frac{-3t}{3})

(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:

\frac{dV(t)}{dt} = 0.6 - 3V(t)

dV(t) = (0.6 - 3V(t))dt

\frac{dV(t)}{0.6 - 3V(t)} = dt

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

(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:
0.2(1-e^{-3t})

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

 

[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. (\frac{dV}{dt}= 0.6- 3V won't work with any of them- as said, it must be 6- 3V.)

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

 

[konichiwa2x]

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

(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
\int uv =u \int v - \int \frac{(du)}{(dx)}\int v 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:

\int \frac{dx}{x} = \ln |x| + C