# Thread: Determining whether acceleration is constant

1. ## Determining whether acceleration is constant

Hi,

I'm trying to solve for the following question:

And the solutions is as follows:

For the solution, I was basically able to reach line 3 until I did not how to simplify the expression any further. What I'm saying is that I don't really understand what the value of "v" within the acceleration equation is suppose to represent and why it is there.

I understand that the acceleration equation is just going to have constants, based on the power rule, but it would be great if I could have some clarification on the steps between lines 3 - 5.

2. ## Re: Determining whether acceleration is constant

$\dfrac{dv}{dt} = \dfrac{1}{2\color{red}{\sqrt{b^2+2gs}}} \times 2g \cdot \color{red}{\dfrac{ds}{dt}}$

note ...

$\color{red}{v = \sqrt{b^2+2gs}}$ and $\color{red}{v = \dfrac{ds}{dt}}$

$a = \dfrac{1}{2\color{red}{v}} \times 2g \cdot \color{red}{v}$

3. ## Re: Determining whether acceleration is constant

So you got to $\displaystyle acc= \frac{dv}{dt}= \frac{dv}{ds}\frac{ds}{dt}= \frac{1}{2}(b^2+ 2gs)^{1/2}\times (2b+ 2g\frac{ds}{dt})$ and want to know how to get the next line, $\displaystyle \frac{1}{2v}(2gv)$?

You are told that $\displaystyle v= \sqrt{b^2+ 2gs}= (b^2+ 2gs)^{1/2}$ so that $\displaystyle (b^2+ 2gs)^{-1/2}= \frac{1}{(b^2+ 2gs)^{1/2}}= \frac{1}{v}$.

You also say "What I'm saying is that I don't really understand what the value of "v" within the acceleration equation is suppose to represent and why it is there." Well, again, you are told that v is the velocity.

You are expected to know that, just as acceleration is the derivative of velocity with respect to time, velocity is the derivative of position with respect to time. That is why $\displaystyle \frac{ds}{dt}$ is replaced with "v".

4. ## Re: Determining whether acceleration is constant

Okay thanks for responding. Your response generally makes sense.

I get why "ds/dt" represents the derivative function that is equal to velocity, however, why would you replace the variable "s" with "ds/dt"? I know that based on the chain rule, that an operation needs to be done on the variable "s", but I don't particularly understand why "ds/dt" is what you replace it with.

Does it follow the same principal when we do implicit differentiation? Like if there is a variable "y" within the function, that the derivative of "y" would be "dy/dx". I'm pretty sure that you follow the same principal, but I just want to make sure.

5. ## Re: Determining whether acceleration is constant

Originally Posted by otownsend
Okay thanks for responding. Your response generally makes sense.

I get why "ds/dt" represents the derivative function that is equal to velocity, however, why would you replace the variable "s" with "ds/dt"? I know that based on the chain rule, that an operation needs to be done on the variable "s", but I don't particularly understand why "ds/dt" is what you replace it with.

Does it follow the same principal when we do implicit differentiation? Like if there is a variable "y" within the function, that the derivative of "y" would be "dy/dx". I'm pretty sure that you follow the same principal, but I just want to make sure.
$s$ is a generic letter used for position (I do not know where this convention came from, but it has been used for many years). Velocity is defined as the instantaneous rate of change in position with respect to time. $\dfrac{ds}{dt}$ means the derivative of $s$ (position) with respect to $t$ (time). That derivative is the instantaneous rate of change in position with respect to time (the definition of velocity). We do not replace $s$ with $\dfrac{ds}{dt}$. We take the derivative of position with respect to time when we want velocity. We keep $s$ as position when we refer to the position. So, if you are told that $v = \sqrt{b^2+2gs}$ where $b,g$ are constants, then you are being told that $\dfrac{ds}{dt} = \sqrt{b^2+2gs}$. This essentially was an implicit differentiation, as the equation relating $s$ and $t$ must have included a function of $s$. In fact, the original formula would have been $\dfrac{\sqrt{b^2+2gs}}{g} = t+C$ where $C$ is some constant we cannot solve for without additional information (I solved this using more advanced mathematics. I do not expect you to understand how. When you get to differential equations, it will become clear.).

Solving for $s$, we get:

$s(t) = \dfrac{g^2(t+C)^2-b^2}{2g} = \dfrac{g}{2}(t+C)^2 -\dfrac{b^2}{2g}$

Taking the derivative with respect to time, we get:

$s'(t) = \dfrac{ds}{dt} = v(t) = g(t+C)$

And finally:

$s''(t) =\dfrac{d^2s}{dt^2} = v'(t) = \dfrac{dv}{dt} = a(t) = g$