Challenging Find maximum deceleration and speed (not max)

[mia2]

Homework Statement

Given $$\frac{d^2s}{dt^2}=-Ae^{s/B}\biggl(\frac{ds}{dt}\biggr)^2$$ Show that maximum deceleration (where $A,B$ are constants, $v=v_0$, and $s=-\infty$) is $\frac{v_0^2}{2eB}$.

Use the substitution $v=\frac{ds}{dt}$.

Homework Equations

See above.

The Attempt at a Solution

Using the substitution, I get $\frac{d^2s}{dt^2}=-Ae^{s/B}(v)^2$. Furthermore, using $v=v_0$ and $s=-\infty$ and using the original function, I get $v=e^{-ABe^{s/B}}v_0$. Combining these two equations, I get
$$\frac{d^2s}{dt^2}=-Ae^{(s/B)-(2ABe^{s/B})}v_0^2$$.
With $s=-\infty$, I end up with:
$$\frac{-Av_0^2}{e^{2AB}}$$, which isn't what's given in the problem. Not sure what I'm doing wrong.

 

[SteamKing]

Your Latex came out garbled. Try hitting the Preview button before you post to make sure everything formats properly.

 

[mia2]

Your Latex came out garbled. Try hitting the Preview button before you post to make sure everything formats properly.

Ok, I've fixed it now. Do you have any suggestions for my physics problem?

 

[Saitama]

$$\frac{d^2s}{dt^2}=-Ae^{(s/B)-(2ABe^{s/B})}v_0^2$$.
With $s=-\infty$, I end up with:
$$\frac{-Av_0^2}{e^{2AB}}$$, which isn't what's given in the problem. Not sure what I'm doing wrong.

Why do you think acceleration is maximum at $s=-\infty$? I don't see how you can find a maximum only by inspection. You will have to differentiate the acceleration function and set the derivative to zero.

 

[HalfLostAndSearching]

I would first check the calculations and make sure all the substitutions and equations have been used correctly. I would also double check the given maximum deceleration value to make sure it is accurate. If it is, then there may be a mistake in the original function or in the given values for A and B. I would suggest checking these values and possibly consulting with a colleague or professor for further clarification. Additionally, I would recommend checking any units used in the calculations to ensure they are consistent and correct.