Question

A particle moves along a straight line with an acceleration a=2v^1/2. If s = 0, v = 9 m/s when t = 0, determine the time for the particle to achieve a velocity of 18 m/s. Also, find the displacement of particle when t = 6s.

Answer 1

Remember \[a = \frac{ dv }{ dt }\] You were given an expression for a in terms of v. So, you have a differential equation in terms of v. This one can be solved for v by the separation of variables method.

Answer 2

\[\int\limits_{9}^{18}dv = \int\limits_{0}^{t}2v ^{\frac{ 1 }{ 2 }}\]

Answer 3

You're missing dt. Also you must gather the v factors to multiply dv, and all the t factors (there aren't any in this problem) to multiply dt, before you integrate.

You wouldn't be able to integrate v^1/2 dt because it is a function of t, and you don't know what that function is yet!

Answer 4

Not sure what you mean by gathering all the v and t factors...

Answer 5

\[\int\limits f(v)dv = \int\limits f(t)dt\]
no v should appear on the right side of that equation.

Answer 6

I'm confused. Is my upper and lower bounds incorrect? Or what?

Answer 7

I mean alright... I'll change it to fvdv=ftdt but not quite sure where you're getting at.

Answer 8

The velocity changes with time. The goal of solving this differential equation is to find v(t). If you didn't bring the v^(1/2) over to the left side you'd have

\[\int\limits dv = 2\int(v(t))^{1/2} dt\]
But you won't be able to integrate that, because you don't know what v(t) is until after you've solved the differential equation.

Answer 9

Oh I see so I have to find v(t):
\[\frac{ dv }{ dt } = 2v ^{\frac{ 1 }{ 2 }}\]
\[\frac{ dv }{v ^{\frac{ 1 }{ 2 }} } = 2dt\]
But now I have v^1/2 and what the heck happens to that?

Answer 10

Nevermind. LOL. So continuing on and ignoring my last question:
\[\int\limits_{9}^{v}v ^{-\frac{ 1 }{2 }}dv = 2\int\limits_{0}^{t}dt\]

Answer 11

\[2v ^{\frac{ 1 }{ 2 }}- 6 = 2t\]

Answer 12

\[v(t)=(\frac{ 2t+6 }{ 2 })^{2}\]

Answer 13

Is that right so far?

Answer 14

It looks right to me.

Answer 15

\[\int\limits dv = 2\int(v(t))^{1/2} dt\]
\[\int\limits_{9}^{18}dv=2\int\limits_{0}^{t}\frac{ 2t+6 }{ 2 }dt\]
\[s(t) = t ^{2} + 6t - 9\]
\[s(6) = 6 ^{2} + 6*6 - 9\]
\[s(6) = 63m\]

Is that right?

Answer 16

No, it's very wrong. Look at your expression for v(t). Should the displacement s(t) be a polynomial of the same order as v(t)? For example, if v(t) = at, what would s(t) be?

\[s(t) = s(0) + \int\limits_{0}^{t}v(\tau)d\tau\]

Answer 17

A problem with what you did is you seem to have thought you would use the original \[a(v) = dv/dt \]
equation to solve for s(t) as if it was the same thing as v, but
\[v(t) = \frac{ ds }{dt }\]

Answer 18

Now I'm confused. Damnit this is too frustrating.

Answer 19

You said this before:
\[\int\limits dv = 2\int(v(t))^{1/2} dt\]
So I went and plugged in v(t). But now you're saying I can't do that?

Answer 20

Can you just show me how to get this done please?

Answer 21

Starting with \[\frac{ dv }{ dt } =2\sqrt{v}\]
you rearranged to get
\[\int\limits \frac{ dv }{\sqrt{v}} =2\int dt \]
You solved this equation correctly for v(t). Now you are done with the equations above, you don't need them anymore except maybe to check your answer is correct. You don't substitute v(t) back into these equations if you want to get s(t)! To get s(t) you use a different equation. The position at time t is equal to the initial position plus the time integral of velocity, and this is always the case regardless of what v(t) is.

Answer 22

But what is v? I know the derivative of position is the velocity but I don't have an equation for the position, I only have the equation of acceleration.

Answer 23

Figured it out. Answer for a is 1.24 seconds and b) I use vdt = ds and use my v(t) for v to solve for the displacement which is 234m.

Answer 24

Anyways, I'll still give you best response since you've helped me in a way.