• rvc

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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  • rvc

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

Physics
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question a) the velocity is given by the subsequent formula: \[\Large v = at\]
  • rvc
is it the differential equation?
please wait

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Other answers:

yes! I think that we have to write a differential equation
  • rvc
yes but im confused which to use
by definition of acceleration, we can write this: \[\Large a = \frac{{dv}}{{dt}} = \frac{8}{{{v^2} + 1}}\]
  • rvc
brb
the space x(t) traveled is given by this differential equation: \[\Large \frac{{dx}}{{dt}} = v\left( t \right)\]
we 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
the question doesn't tell us about the shape
yes 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
sincerely I don't know how to overcome this difficulty
  • rvc
we don't have t in the equation
please wait I ask to another helper
@mathstudent55 please help
@e.mccormick please help
  • rvc
I think we have these formulas: \[\Large \rm a=v \frac{ dv }{ dx }\] \[\Large \rm v=\frac{ dx }{ dt }\] \[\Large \rm a=\frac{ dv }{ dt }\]
yes! nice idea!
it 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.
we have: \[\Large \frac{{dx}}{{dv}} = \frac{v}{a} = \frac{{v\left( {{v^2} + 1} \right)}}{8}\]
  • rvc
Okay @Haseeb96 Thanks bro :)
thanks @Haseeb96 actually @rvc helped me with her nice idea
ok! 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
wait how dx/dv? can we write like that way?
since: \[\Large \frac{{dx}}{{dv}} = \frac{1}{{\frac{{dv}}{{dx}}}}\]
  • rvc
oh okay
after a substitution, we gave to solve a qudratic equation for v^2
we have*
for part b) we can use the same formula: \[\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)\]
since the motion is the same. We have only to substitute v=64.8 at the right side
  • rvc
and the first condition?
do you mean part a) ?
  • rvc
yep
we have to substitute your data, so we can write this: \[\Large \begin{gathered} 20 = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right) \hfill \\ \hfill \\ 20 = \frac{1}{8}\left( {\frac{{{v^4} + 2{v^2}}}{4}} \right) \hfill \\ \hfill \\ 640 = {v^4} + 2{v^2} \hfill \\ \hfill \\ {v^4} + 2{v^2} - 640 = 0 \hfill \\ \end{gathered} \]
now, we can make this change of variable: w=v^2, so we get: \[\Large {w^2} + 2w - 640 = 0\]
  • rvc
wait 8 X 20 = 160
  • rvc
oh okayyy
now, 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
I got this: \[\Large w = {v^2} = - 1 + \sqrt {640} \cong - 1 + 25.3 = 24.3\]
and 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
i read every step i understand
ok! :)

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