Challenging Find maximum deceleration and speed (not max)

AI Thread Summary
The discussion focuses on determining the maximum deceleration from the equation given, where the second derivative of position with respect to time is defined. The user attempts to derive the maximum deceleration using a substitution for velocity and evaluates the conditions at s approaching negative infinity. They express confusion over their calculations, which do not yield the expected result of v_0^2/(2eB). Another participant suggests that to find the maximum deceleration, one must differentiate the acceleration function and set the derivative to zero, rather than relying on inspection. The conversation emphasizes the importance of proper differentiation in solving the problem accurately.
mia2
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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.
 
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Your Latex came out garbled. Try hitting the Preview button before you post to make sure everything formats properly.
 
SteamKing said:
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?
 
mia2 said:
$$\frac{d^2s}{dt^2}=-Ae^{(s/B)-(2ABe^{s/B})}v_0^2$$.
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.
 
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