Related Rates and acceleration

[Mspike6]

I have a question about a related rates, i already solved it, but i would like someone to confirm it to me .

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

Solution:
\frac{d}{dt} (V) = \frac{d}{dt}[\frac{150s}{3+s}]\frac{dv}{dt} = 150 \frac{d}{dt} \frac{s}{3+s}\frac{dv}{dt} =150 [ \frac{(3+s)(s') - (s)(s')}{(3+s)^2]}\frac{dv}{dt} = \frac{150s'}{(3+s)^2}a= \frac{450v}{(3+s)^2}

As you can see my final answer is a in terms of v and s, not s only .. .but am not sure what to do to getrid of this v..

Thanks

 

[ƒ(x)]

There's a mistake between the third and forth lines. Also, what substitution can you make to get rid of v?

 

[Mspike6]

There's a mistake between the third and forth lines. Also, what substitution can you make to get rid of v?

thanks, the 4th line should be \frac{dv}{dt}= \frac{450v}{(3+s)^2}

am not sure what sub. get rid of v.
but v' will give me a instead of v, but is that possible, i mean, if i want to get V' i will have to take thderivative for the whole thing.

 

[kreil]

Your original velocity equation expresses v in terms of s; use that.

 

[Mspike6]

Your original velocity equation expresses v in terms of s; use that.

sorry but i don't know what you mean .

The Question asked me to find a in terms of s , not v in terms of s

 

[kreil]

your expression for a is correct, but it contains a 'v'. in order to express a completely in terms of s, you need to use the equation v=\frac{150s}{3+s} and plug this into your equation for a.

 

[Mspike6]

your expression for a is correct, but it contains a 'v'. in order to express a completely in terms of s, you need to use the equation v=\frac{150s}{3+s} and plug this into your equation for a.

Ah, that make sense, thank you :D