How can I solve this integrals problem?

[lasvegas666]

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[mario99]

View attachment 37812'

Integrate both sides as before.

 

[lasvegas666]

Integrate both sides as before.

So velocity is a derivative of speed?

I got it now.

 

[mario99]

So velocity is a derivative of speed?

I got it now.

Velocity $v(t)$ is the derivative of position $s(t)$.

 

[The Highlander]

Velocity is not the derivative of "speed" or "position"; it is the derivative of displacement wrt time: $\displaystyle \frac{\mathrm{ds}}{\mathrm{dt}}$

 

[mario99]

Velocity is not the derivative of "speed" or "position"; it is the derivative of displacement wrt time: $\displaystyle \frac{\mathrm{ds}}{\mathrm{dt}}$

You said velocity is the derivative of displacement$\displaystyle \frac{d}{dt}(s - s_0)$, yet, you wrote position $\displaystyle \frac{d}{dt}(s)$.

 

[The Highlander]

You said velocity is the derivative of displacement$\displaystyle \frac{d}{dt}(s - s_0)$, yet, you wrote position $\displaystyle \frac{d}{dt}(s)$.

Displacement (s) is a vector quantity (with magnitude & direction), it is not (s - s₀).

It's magnitude might be determined by (p - p₀), where p & p₀ are two positions (current & initial respectively) and so it should be written as:-

$\displaystyle \frac{\mathrm{ds}}{\mathrm{dt}}$ or $\displaystyle \frac{\mathrm{d}}{\mathrm{dt}}\mathrm{s}$​

 

[mario99]

Displacement (s) is a vector quantity (with magnitude & direction), it is not (s - s₀).

It's magnitude might be determined by (p - p₀), where p & p₀ are two positions (current & initial respectively) and so it should be written as:-

$\displaystyle \frac{\mathrm{ds}}{\mathrm{dt}}$ or $\displaystyle \frac{\mathrm{d}}{\mathrm{dt}}\mathrm{s}$​

You meant to say $\bold{v} = \displaystyle \frac{d}{dt}\bold{s} = \displaystyle \frac{d}{dt}(\bold{p} - \bold{p_0})$

But

$\bold{v} = \displaystyle \frac{d}{dt}(\bold{p} - 0) = \displaystyle \frac{d}{dt}(\bold{p})$

Which is still the derivative of position.

 

[The Highlander]

You meant to say $\bold{v} = \displaystyle \frac{d}{dt}\bold{s} = \displaystyle \frac{d}{dt}(\bold{p} - \bold{p_0})$

But

$\bold{v} = \displaystyle \frac{d}{dt}(\bold{p} - 0) = \displaystyle \frac{d}{dt}(\bold{p})$

Which is still the derivative of position.

No, it is the derivative of displacement.
Please don't engage in any further semantic argument; this is an already well defined concept.

 

[khansaheb]

You have been asked to find the position function - given the speed function

 

[mario99]

No, it is the derivative of displacement.
Please don't engage in any further semantic argument; this is an already well defined concept.

I have just shown you that it is not wrong to say velocity is the derivative of position in this problem of #1. It is up to you how far you wanna go with your argument.

I am done.