Relation between distance s and velocity v is given by $v=\dfrac {150s} {3+s}$

Question

I am stuck on this related rates question:

The relation between distance $s$ and velocity $v$ is given by $v=\dfrac {150s} {3+s}$. Find the acceleration in terms of s.

So far I have:

$$\dfrac {dv} {dt}=\dfrac {\left( 3+s\right) \dfrac {d} {dt}\left( 150s\right) -150s\dfrac {d} {d t}\left( 3+s\right) } {\left( 3+s\right) ^{2}}$$

$$\dfrac {dv} {dt}=\dfrac {\left( 3+s\right) \left( 150\right)\dfrac {ds} {dt}-150s\left( 1\right) \dfrac {ds} {dt}} {\left( 3+s\right) ^{2}}$$

Let $a=\dfrac{dv}{dt}$ and $v=\dfrac{ds}{dt}$

$$a=\dfrac {450v+150sv-150sv} {\left( 3+s\right) ^{2}}$$ $$a=\dfrac {450v} {\left( 3+s\right) ^{2}}[m/s^2]$$

Is my solution correct? Am I close?

Answer 1

The relation between distance s and velocity v is given by v=150s/(3+s). Find the acceleration in terms of s.

In order to find acceleration, a = dv/dt

$$v = 150s/(3+s)$$

Using the Quotient Rule: $$dv/dt = ((150s)'*(3+s)-150s*(3+s)')/(3+s)^2$$

$$dv/dt = (150s'*(3+s)-150s*(s)')/(3+s)^2$$

and s'=v so:

$$dv/dt = (150v*(3+s)-150s*v)/(3+s)^2$$

$$dv/dt = (450v+150s*v-150s*v)/(3+s)^2$$

$$dv/dt = (450v)/(3+s)^2$$

So yeah looks like you are right.

Answer 2

here is another way to do this. $$ v = \frac{150s}{3+s} = 150 - \frac{450}{s+3}$$

now use chain rule $$\frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = \frac{450}{(s+3)^2} v = \frac{450*150s}{(s+3)^3}$$