fiddling with differentials

[shakalandro]

given the function a(v) = -uv[sup:1tiuyogg]2[/sup:1tiuyogg] and the property that a(t) = dv/dt = v(dv/dx) I was supposed to find dv/dx which of course = -uv,
next I had to solve the differential equation and determine what u was given that v(0) = 15 and v(2000/5280) = 15,
I arrived at V = (66/25)(ln.1)e[sup:1tiuyogg]-6.078x[/sup:1tiuyogg]. So finally, the last part of the question which I cannot figure out how to do is to find the time it takes for an object under this model to slow down to 15mph?

I think this a matter of deriving another diff eq, but I can't determine what it is.

 

[Deleted member 4993]

given the function a(v) = -uv[sup:2uw8jlen]2[/sup:2uw8jlen] and the property that a(t) = dv/dt = v(dv/dx) I was supposed to find dv/dx which of course = -uv,
next I had to solve the differential equation and determine what u was given that v(0) = 15 and v(2000/5280) = 15,
I arrived at V = (66/25)(ln.1)e[sup:2uw8jlen]-6.078x[/sup:2uw8jlen]. So finally, the last part of the question which I cannot figure out how to do is to find the time it takes for an object under this model to slow down to 15mph?

I think this a matter of deriving another diff eq, but I can't determine what it is.

When you say V(0) = 15 is that at t=0 or x=0

you say

V = (66/25)(ln.1)e[sup:2uw8jlen]-6.078x[/sup:2uw8jlen].

however

ln(1) = 0

So I am not sure what you are saying!!

Please show your work, so that we know where to begin to help you.

 

[fasteddie65]

given the function a(v) = -uv[sup:2qoffyaz]2[/sup:2qoffyaz] and the property that a(t) = dv/dt = v(dv/dx) I was supposed to find dv/dx which of course = -uv,

The derivative of -uv^2 is -2uv, if you are differentiating with respect to v.

The derivative of -uv^2, using the Product Rule, is -v^2 du/dt - 2uv dv/dt, if you are differentiating with respect to t.

 

[shakalandro]

sorry, it's v(0) = 150 and this is for v(x), though it shouldn't really matter because v(t) at t = 0 is the same as v(x) at x = 0.

Edit! here is my work for parts a and b

a(v) = -uv[sup:e7vpopqb]2[/sup:e7vpopqb]= a(t) = v(dv/dx)

dv/dx= -uv

solving the ODE:

v(x) = ce[sup:e7vpopqb]-ux[/sup:e7vpopqb], v(0) = 150
c = 150
v(x) = 150e[sup:e7vpopqb]-ux[/sup:e7vpopqb], v(2000/5280) = 15
u = 6.078
v(x) = 150e[sup:e7vpopqb]-6.078x[/sup:e7vpopqb]

so now what I need is a formula for v(t) that I can use to find the time where v = 15mph

 

[shakalandro]

Given that a(t) = v(dv/dx), can this be integrated to produce v(t), I'm just not sure how to deal with such an integral

 

[Deleted member 4993]

Given that a(t) = v(dv/dx), can this be integrated to produce v(t), I'm just not sure how to deal with such an integral

I think you ought to start with

a(v) = dv/dt

 

[shakalandro]

I don't understand, wouldn't a(t) = dv/dt, not a(v)

 

[Deleted member 4993]

I don't understand, wouldn't a(t) = dv/dt, not a(v)

a(v) = -uv[sup:xrxku513]2[/sup:xrxku513]

dv/dt = -uv[sup:xrxku513]2[/sup:xrxku513]

dv/v[sup:xrxku513]2[/sup:xrxku513] = -u dt

and so on .....

 

[shakalandro]

again, I do not see how acceleration in terms of velocity is the derivative of velocity with respect to time, I have only ever seen it where acceleration in terms of time is the derivative of velocity with respect to time

 

[Deleted member 4993]

again, I do not see how acceleration in terms of velocity is the derivative of velocity with respect to time (that is the physical definition of acceleration), I have only ever seen it where acceleration in terms of time is the derivative of velocity with respect to time

 

[shakalandro]

again, I do not see how acceleration in terms of velocity is the derivative of velocity with respect to time (that is the physical definition of acceleration), I have only ever seen it where acceleration in terms of time is the derivative of velocity with respect to time

yes that is the physical definition of acceleration in terms of time, not in terms of velocity, which is what we are dealing with