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I've problem with formulating the following problem. I guess I need to express $v(d)$ in $v(t)$ but since $d=v*t$ I can't just replace $d$ with $v*t$ since I would get $v(t) = v...$, a recursive function.

A particle moves in a straight line. The velocity of the particle ($v$) depends on the traveled distance ($d) according to:

$$ v = \frac{3d+4}{2d+1} $$

where $d$ is the distance from the particle to its starting point. Calculate the particle acceleration when d = 2. (The derivate of speed depending on time gives the acceleration).

iveqy
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3 Answers3

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$d$ is a function $d(t)$ of time, and by extension $v(t) = d'(t)$ is as well. You have an expression for the first derivative $d'(t) = v(t)$, the text asks you to find the second derivative at the time when $d(t) = 2$. So, we need to differentiate with respect to $t$: $$ a(t) = v'(t) = \frac{ 3(2d(t) + 1)-2(3d(t) + 4)}{(2d(t) + 1)^2}\cdot d'(t)\\ = \frac{-5}{(2d(t) + 1)^2}\cdot v(t) = \frac{-5(3d(t) + 4)}{(2d(t) + 1)^3} $$ entering $d(t) = 2$, we get $a(t) = -2/5$

Arthur
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  • shouldn't that be $a(t) = -\frac{2}{5}$? – iveqy Oct 23 '13 at 12:13
  • @iveqy I fail to see where the negative sign comes from. – Arthur Oct 23 '13 at 12:18
  • I found it. When you do the derivate I guess you switched the functions. let $f(x) = 3x+4$ and $g(x) = 2x+1$ then $\frac{f(x)}{g(x)} = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$. Would you agree? If so, please let me know so that I/you can edit your answer so I can approve it. Many thanks for your helpfullness (on this an other questions)! – iveqy Oct 24 '13 at 09:22
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    @iveqy You were right, I had switched the functions. I have changed it now. – Arthur Oct 24 '13 at 10:23
  • but it shouldn't be $2(3d+4)−3(2d+1)$, it should be $3(2d+1)-2(3d+4)$ – iveqy Oct 24 '13 at 10:28
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Hint, we have $a(t)=\frac d{dt}v(t)=\frac d{dt}\frac {3d+4}{2d+1}$. Now apply the chain rule, noting that $\frac {dd}{dt}=v(t)$, then evaluate at $d=2$ and report the result.

Ross Millikan
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The equation $d=v\cdot t$ is only true for constant velocities. If $v$ is not constant, it changes to $\frac{\mathrm dd}{\mathrm dt} = v$ which just says that the velocity is the time derivative of the distance travelled. You are looking for the acceleration $a=\frac{\mathrm dv}{\mathrm dt}$, applying the chain rule gives $a=\frac{\mathrm dv}{\mathrm dd}\cdot\frac{\mathrm dd}{\mathrm dt}=\frac{\mathrm dv}{\mathrm dd}\cdot v$.

Christoph
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