Kinematics Matching Initial Conditions

Question

The acceleration of a body is given in terms of the displacement $s$ meters as $$a = \frac{2s}{s^2+1}$$ Give a formula for the velocity as a function of displacement given that when $s=1,v = 2.$

$$\therefore \frac{1}{2}v^2 = ln(s^2+1)$$

$$v = \sqrt{2ln(s^2+1)}$$

To match initial conditions, I did the following:

$$v = \sqrt{2ln(s^2+1)} + k$$ $$2 = \sqrt{2ln(2)} + k$$ $$k = 2 - \sqrt{ln(4)}$$ $$v = \sqrt{2ln(s^2+1)} + (2 - \sqrt{ln(4)})$$

However, the answer given adds a constant $k$ when in terms of $\frac{1}{2}v^2$:

Can we say that one answer is correct? This goes on to cause trouble in further subsections of this question; Is the answer provided the standard way of solving such questions? If so, why?

The issue with your solution, compared with the given answer, is that you don't add the constant of integration when you evaluate the integral. The answer is correct because it adds the integration constant when the integral is performed — eepperly16, Aug 07 '16 at 03:55
You can post that as an answer, and I'll give you credit for it. — StopReadingThisUsername, Aug 07 '16 at 04:19

score 1 · Accepted Answer · answered Aug 07 '16 at 19:40

1

You might like to consider the following trick.

When you have separated the variables and applied the integration signs, why don't you also apply the limits as shown below. Then you don't have to worry about the constant of integration at all:

$$\int vdv=\int \frac{2s}{s^2+1} ds\Rightarrow\int_{\color{red}{2}}^{\color{green}{v}} vdv=\int_{\color{red}{1}}^{\color{green}{s}} \frac{2s}{s^2+1} ds$$

answered Aug 07 '16 at 19:40

David Quinn

34,121

Interesting...thanks for this! (+1) – StopReadingThisUsername Aug 08 '16 at 06:11

Kinematics Matching Initial Conditions

1 Answers1