# Homework Help: Motion of a body -calc

1. Sep 1, 2006

### 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?

Last edited: Sep 1, 2006
2. Sep 2, 2006

### Päällikkö

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

3. Sep 2, 2006

### 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,

4. Sep 2, 2006

### Päällikkö

I googled for "separable differential equation" and found a decent looking text:

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:
$$\frac{dV(t)}{dt} = 6 - 3V(t)$$

Last edited by a moderator: Apr 22, 2017
5. Sep 2, 2006

### 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).)

6. Sep 2, 2006

### konichiwa2x

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?

Last edited: Sep 2, 2006
7. Sep 2, 2006

### Päällikkö

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