Question

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

Answer 1

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

Answer 2

is it the differential equation?

Answer 3

please wait

Answer 4

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

Answer 5

yes
but im confused which to use

Answer 6

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

Answer 7

brb

Answer 8

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

Answer 9

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:
\[\Large v = v\left( t \right)\]

Answer 10

the question doesn't tell us about the shape

Answer 11

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

Answer 12

sincerely I don't know how to overcome this difficulty

Answer 13

we don't have t in the equation

Answer 14

please wait I ask to another helper

Answer 15

@mathstudent55 please help

Answer 16

@e.mccormick please help

Answer 17

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

Answer 18

yes! nice idea!

Answer 19

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.

Answer 20

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

Answer 21

Okay @Haseeb96 Thanks bro :)

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

since:
\[\Large \frac{{dx}}{{dv}} = \frac{1}{{\frac{{dv}}{{dx}}}}\]

Answer 26

oh okay

Answer 27

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

Answer 28

we have*

Answer 29

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

Answer 30

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

Answer 31

and the first condition?

Answer 32

do you mean part a) ?

Answer 33

yep

Answer 34

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

Answer 35

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

Answer 36

wait
8 X 20 = 160

Answer 37

oh okayyy

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

i read every step
i understand

Answer 42

ok! :)