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rvc
 one year ago
The acceleration of the train starting from the rest is a= 8/(v^2+1).
Find
a. v=? when x= 20m
b. x=? when v=64.8km/hr
rvc
 one year ago
The acceleration of the train starting from the rest is a= 8/(v^2+1). Find a. v=? when x= 20m b. x=? when v=64.8km/hr

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3question a) the velocity is given by the subsequent formula: \[\Large v = at\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.1is it the differential equation?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3yes! I think that we have to write a differential equation

rvc
 one year ago
Best ResponseYou've already chosen the best response.1yes but im confused which to use

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3by definition of acceleration, we can write this: \[\Large a = \frac{{dv}}{{dt}} = \frac{8}{{{v^2} + 1}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3the space x(t) traveled is given by this differential equation: \[\Large \frac{{dx}}{{dt}} = v\left( t \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have to integrate that ODE, so we have: \[\Large x\left( t \right) = \int {v\left( t \right)dt} \] then we have to know the shape of the function: \[\Large v = v\left( t \right)\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.1the question doesn't tell us about the shape

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3yes I know, we have to compute it, nevertheless it is not simple, since after a simple integration, we get: \[\Large \frac{{{v^3}}}{3} + v = 8t\] as you can easily check

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3sincerely I don't know how to overcome this difficulty

rvc
 one year ago
Best ResponseYou've already chosen the best response.1we don't have t in the equation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3please wait I ask to another helper

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3@mathstudent55 please help

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3@e.mccormick please help

rvc
 one year ago
Best ResponseYou've already chosen the best response.1I think we have these formulas: \[\Large \rm a=v \frac{ dv }{ dx }\] \[\Large \rm v=\frac{ dx }{ dt }\] \[\Large \rm a=\frac{ dv }{ dt }\]

Haseeb96
 one year ago
Best ResponseYou've already chosen the best response.0it is very easy I think @Michele_Laino can figure out your this problem. I have to go on my roof. Because there is some huge storm coming.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have: \[\Large \frac{{dx}}{{dv}} = \frac{v}{a} = \frac{{v\left( {{v^2} + 1} \right)}}{8}\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.1Okay @Haseeb96 Thanks bro :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3thanks @Haseeb96 actually @rvc helped me with her nice idea

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3ok! we have only to integrate that last ODE, so we can write this: \[\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.1wait how dx/dv? can we write like that way?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3since: \[\Large \frac{{dx}}{{dv}} = \frac{1}{{\frac{{dv}}{{dx}}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3after a substitution, we gave to solve a qudratic equation for v^2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3for part b) we can use the same formula: \[\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3since the motion is the same. We have only to substitute v=64.8 at the right side

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3do you mean part a) ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have to substitute your data, so we can write this: \[\Large \begin{gathered} 20 = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right) \hfill \\ \hfill \\ 20 = \frac{1}{8}\left( {\frac{{{v^4} + 2{v^2}}}{4}} \right) \hfill \\ \hfill \\ 640 = {v^4} + 2{v^2} \hfill \\ \hfill \\ {v^4} + 2{v^2}  640 = 0 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, we can make this change of variable: w=v^2, so we get: \[\Large {w^2} + 2w  640 = 0\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, we have to solve with respect to w, and we have to pick the positive solution of w, since we have to remember that w=v^2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I got this: \[\Large w = {v^2} =  1 + \sqrt {640} \cong  1 + 25.3 = 24.3\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3and hence: \[\Large v = \sqrt {24.3} \cong 5.03\] I write the positive solution only for v, since I believe that v is the magnitude of the velocity of our particle

rvc
 one year ago
Best ResponseYou've already chosen the best response.1i read every step i understand
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