## 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

1. Michele_Laino

question a) the velocity is given by the subsequent formula: $\Large v = at$

2. rvc

is it the differential equation?

3. Michele_Laino

4. Michele_Laino

yes! I think that we have to write a differential equation

5. rvc

yes but im confused which to use

6. Michele_Laino

by definition of acceleration, we can write this: $\Large a = \frac{{dv}}{{dt}} = \frac{8}{{{v^2} + 1}}$

7. rvc

brb

8. Michele_Laino

the space x(t) traveled is given by this differential equation: $\Large \frac{{dx}}{{dt}} = v\left( t \right)$

9. Michele_Laino

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: $\Large v = v\left( t \right)$

10. rvc

the question doesn't tell us about the shape

11. Michele_Laino

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

12. Michele_Laino

sincerely I don't know how to overcome this difficulty

13. rvc

we don't have t in the equation

14. Michele_Laino

15. Michele_Laino

16. Michele_Laino

17. rvc

I think we have these formulas: $\Large \rm a=v \frac{ dv }{ dx }$ $\Large \rm v=\frac{ dx }{ dt }$ $\Large \rm a=\frac{ dv }{ dt }$

18. Michele_Laino

yes! nice idea!

19. Haseeb96

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.

20. Michele_Laino

we have: $\Large \frac{{dx}}{{dv}} = \frac{v}{a} = \frac{{v\left( {{v^2} + 1} \right)}}{8}$

21. rvc

Okay @Haseeb96 Thanks bro :)

22. Michele_Laino

thanks @Haseeb96 actually @rvc helped me with her nice idea

23. Michele_Laino

ok! we have only to integrate that last ODE, so we can write this: $\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)$

24. rvc

wait how dx/dv? can we write like that way?

25. Michele_Laino

since: $\Large \frac{{dx}}{{dv}} = \frac{1}{{\frac{{dv}}{{dx}}}}$

26. rvc

oh okay

27. Michele_Laino

after a substitution, we gave to solve a qudratic equation for v^2

28. Michele_Laino

we have*

29. Michele_Laino

for part b) we can use the same formula: $\Large x\left( v \right) = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right)$

30. Michele_Laino

since the motion is the same. We have only to substitute v=64.8 at the right side

31. rvc

and the first condition?

32. Michele_Laino

do you mean part a) ?

33. rvc

yep

34. Michele_Laino

we have to substitute your data, so we can write this: $\Large \begin{gathered} 20 = \frac{1}{8}\left( {\frac{{{v^4}}}{4} + \frac{{{v^2}}}{2}} \right) \hfill \\ \hfill \\ 20 = \frac{1}{8}\left( {\frac{{{v^4} + 2{v^2}}}{4}} \right) \hfill \\ \hfill \\ 640 = {v^4} + 2{v^2} \hfill \\ \hfill \\ {v^4} + 2{v^2} - 640 = 0 \hfill \\ \end{gathered}$

35. Michele_Laino

now, we can make this change of variable: w=v^2, so we get: $\Large {w^2} + 2w - 640 = 0$

36. rvc

wait 8 X 20 = 160

37. rvc

oh okayyy

38. Michele_Laino

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

39. Michele_Laino

I got this: $\Large w = {v^2} = - 1 + \sqrt {640} \cong - 1 + 25.3 = 24.3$

40. Michele_Laino

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

41. rvc

i read every step i understand

42. Michele_Laino

ok! :)