# Integral with initial conditions

#### Karol

1. The problem statement, all variables and given/known data

2. Relevant equations
$$\frac{dy}{dx}=f(x)~\rightarrow~dy=f(x)dx~\rightarrow~y=\int f(x)dx$$
$$F=ma,~a=\frac{dv}{dt}$$

3. The attempt at a solution
$$-\frac{mgR}{s^2}=mv\left( \frac{dv}{ds} \right)\rightarrow~v^2=\frac{2gR}{s}+C$$
$$v_0(s=R)=\sqrt{2gR}\rightarrow~C=2g(R-1)$$
$$v^2=\frac{2gR}{s}+2g(R-1)$$
Not good

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#### haruspex

Homework Helper
Gold Member
2018 Award
The link to the image for the start of the question does not work for me.
Your very first equation looks dimensionally wrong. If s is a displacement then gR/s2 has dimension T-2.

Edit. On reloading the page the first link works, and I see that the incorrect equation is provided. It should be mgR2/s2.

Last edited:

#### Karol

$$v=\frac{ds}{dt}=v_0\sqrt{2gR}\rightarrow~s=\int v_0\sqrt{2gR} \,dt$$
But that is a function of s, not t. nowhere there is t, nor in $~a=v\frac{dv}{ds}$ nor in $~v=v_0\sqrt{2gR}$

#### haruspex

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Gold Member
2018 Award
$$v=\frac{ds}{dt}=v_0\sqrt{2gR}\rightarrow~s=\int v_0\sqrt{2gR} \,dt$$
But that is a function of s, not t. nowhere there is t, nor in $~a=v\frac{dv}{ds}$ nor in $~v=v_0\sqrt{2gR}$
I have no idea what you are doing there. v is not v0√(2gR). Where did that come from?

#### Karol

$$\left\{ \begin{array}{l} v^2=\frac{2gR^2}{s}+C \\ v_0(s=R)=\sqrt{2gR} \end{array} \right. \rightarrow~C=0$$
$$v^2=\frac{2gR^2}{s} \rightarrow~v=R\sqrt{\frac{2g}{s}}=v_0\sqrt{\frac{R}{s}}$$

#### haruspex

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Gold Member
2018 Award
$$\left\{ \begin{array}{l} v^2=\frac{2gR^2}{s}+C \\ v_0(s=R)=\sqrt{2gR} \end{array} \right. \rightarrow~C=0$$
$$v^2=\frac{2gR^2}{s} \rightarrow~v=R\sqrt{\frac{2g}{s}}=v_0\sqrt{\frac{R}{s}}$$
Yes.

#### Karol

But that is a function of s, not t. nowhere there is t, nor in $~a=v\frac{dv}{ds}$ nor in $~v=v_0\sqrt{\frac{R}{s}}$

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#### Karol

$$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$

#### ehild

Homework Helper
$$v=\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~s=\int v_0\sqrt{\frac{R}{s}} \,dt$$
It does not go anywhere. You have a separable differential equation, you can integral a function of s with respect to s. and a function of t with respect to t.

#### Karol

$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\sqrt{s}\,ds=v_0\sqrt{R}\,dt$$
$$s\sqrt{s}=v_0\sqrt{R}\cdot t \rightarrow~\sqrt{s^3}=v_0\sqrt{R}\cdot t$$
It isn't $~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]$

#### ehild

Homework Helper
$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\sqrt{s}\,ds=v_0\sqrt{R}\,dt$$
$$s\sqrt{s}=v_0\sqrt{R}\cdot t \rightarrow~\sqrt{s^3}=v_0\sqrt{R}\cdot t$$
It isn't $~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]$
Did you forget how to integrate a power function? And you also omitted the integration constant.

#### Karol

$$\frac{ds}{dt}=v_0\sqrt{\frac{R}{s}}\rightarrow~\int \sqrt{s}\,ds=\int v_0\sqrt{R}\,dt$$
$$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+C$$
$$s_0(t=0)=R\rightarrow~C=\frac{2}{3}R^{3/2}$$
$$\frac{2}{3}s^{3/2}=v_0\sqrt{R}\cdot t+\frac{2}{3}R^{3/2}$$
$$\rightarrow~~\sqrt{s^3}=R^{3/2}[ 1+(3v_0t/2R) ]$$

#### ehild

Homework Helper
Correct, at last.

#### Karol

Thank you ehild and Haruspex

"Integral with initial conditions"

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