2

A particle moves with the velocity given by:

$$v(s(t)) = \frac{3s(t) + 4}{2s(t)+1}$$ where s(t) is the distance traveled.

Find the acceleration when s(t) = 2.

My attempt: $$a(s(t)) = \frac{\mathrm d}{\mathrm dV(s(t))}\left(\frac{3s(t) + 4}{2s(t)+1}\right) = \frac{(3s(t)+4)2 - 3(2s(t)+1)}{(2s(t)+1)^2}$$ $$a(2) = \frac{10\times 2-3\times 5}{25} = \frac{5}{25} = \frac{1}{5}.$$

The answer is however supposed to be $-\frac{2}{5}$. Why?

E. Joseph
  • 14,843
I_snort_TNT
  • 101
  • Aside from the sign error, you’re computing $dv/ds$ instead of $dv/dt$, which per the chain rule requires a factor of $ds/dt$. – amd Dec 19 '16 at 21:24

3 Answers3

3

The method is right, but you are wrong about your derivate.

You have

$$\frac{\mathrm d}{\mathrm d t}\left(\frac fg\right)=\frac{f'g-g'f}{g^2}$$

and not

$$\frac{\mathrm d}{\mathrm d t}\left(\frac fg\right)=\frac{fg'-f'g}{g^2}.$$

So the answer would be

$$-\frac 15.$$

E. Joseph
  • 14,843
  • Thank you, I make such silly mistakes. I would also answer $-\frac{1}{5}$ but the answer should apparently be $-\frac{2}{5}$. Any idea why? – I_snort_TNT Dec 19 '16 at 15:17
  • @I_snort_TNT I have no idea why the answer would be $-2/5$. May be there is a mistake in the correction. – E. Joseph Dec 19 '16 at 15:18
  • You’re both computing $dv/ds$ instead of $dv/dt$, which requires another factor of $ds/dt$ per the chain rule. – amd Dec 19 '16 at 21:21
2

you have $$v = \frac{3s+4}{2s+1}= \frac 32 + \frac 5{2(2s+1)}, s = 2, v = 2.$$differencing you get $$ dv = -\frac{5ds}{(2s+1)^2}=-\frac{ds}{5} \mbox{ at }s = 2.$$ dividing by $dt$ and using $\frac {ds}{dt} = v = 2,$ you get $\frac{dv}{dt} = -\frac25 \mbox{ at } s = 2, v = 2. $

abel
  • 29,170
0

You have a function $v$ that gives the velocity $ds/dt$ in terms of position and an unknown function $s$ that gives position in terms of time. So the function that you are differentiating is really $v\circ s$. By the chain rule, $${dv\over dt}(t)={dv\over ds}(s(t))\cdot{ds\over dt}(t)={dv\over ds}(s(t))\cdot(v\circ s)(t)={3(2s(t)+1)-2(3s(t)+4)\over(2s(t)+1)^2}\cdot{3s(t)+4\over2s(t)+1}.$$ Plugging in the known value of $2$ for $s(t)$ produces $-\frac25$ as required.

amd
  • 53,693