Acceleration depending on time

By Anonymous on Thursday, November 23, 2000 - 12:39 am :

Hi,

I doing some work on "Acceleration depending on time". I cannot do part (b) of the following question. Can anyone help out? It would be helpful to me if you could also show your steps and a small commentary to help my understanding. Thanks.

A particle P moves along the positive x-axis. Its acceleration is (x + 3)ms^-2 when its displacement from the origin O is x metres. Given that initially, when t=0, the velocity of P is 3ms^-1 in the direction of Ox and x=0 , obtain
(a) the speed v ms^-1 of P as a function of x
(b) x as a function of t

(a)
here's what I did:

0.5v²=ò(x+3) dx+A
if x=0 and v=3
after evaluation, v = x + 3

(b)
not sure of what do here

Thanks in advance for any help.

By James Lingard (Jchl2) on Thursday, November 23, 2000 - 01:29 am :

Hi,

First, your answer to part (a) is correct, and it looks like you're going about it the right way. Here's how I did it:

KE gained by particle = work done on particle

mv²/2 - mu²/2 = òF.dy = òm a.dy

v² - u² = 2.òa.dy

v² - 3² = 2.ò(x + 3).dy = 2[y² /2 + 3y], evaluated between y = x and y = 0

v² - 3² = 2(x² /2 + 3x) = x² + 6x

v² = x² + 6x + 9 = (x + 3)²

v = ±(x + 3)

But you know to take the +ve square root because v = 3 (not -3) when x = 0.

For part (b), we have

dv/dt = a = x + 3 = v

and so we have the differential equation

dv/dt - v = 0

with solutions

v = A e^t

Then v = x + 3 and so x = v - 3, so

x = A e^t - 3

but since x = 0 when t = 0, 0 = A - 3, so A = 3 and

x = 3e^t - 3

I hope that was clear enough. If not, please let me know.

If you couldn't do that then please be reassured by the fact that it took me about half an hour to realise how to do that question (it's been quite a long time since I did any of these sort of questions).

James.

By Susan Langley (Sml30) on Thursday, November 23, 2000 - 09:03 am :

Another way to do (b) is:

You have v=dx/dt=x + 3

So ò1/(x + 3) dx = ò1 dt

so ln(x+3)=t+c

x=e^t+c -3

x=A e^t -3

and apply the initial conditions like James has.

I don't know whether you have done enough calculus for this, but if you have then I think this is generally a nicer way of doing it. You can often apply what you had in a previous part of the problem to later parts like this. When you can it tends to be faster.

Hope that gives you some more insight.

Susan

By Anonymous on Thursday, November 23, 2000 - 12:50 pm :

Cool! - 2 methods to get to the solution.
Thanks James and Susan! Your help very much appreciated.

By Anonymous on Saturday, November 25, 2000 - 08:34 pm :

I can't figure out part (b), can anyone help.

A particle P of mass 1.5kg moves in a straight line through a fixed point O. At time t sec after passing through O the distance of P from O is x cm and the force acting on P has magnitude (3x + 6)N and directed away from O. Given that P passes through O with speed 2(2)^0.5 ms^-1 calculate, (a) speed of P then x=5, (b) the value of t when x=20

(a) f=m a, ò3x+6 dx=0.5m v²Þ (3/2)x²+6x+c=(1/2)(3/2)v², but c=6, when v=2(2)^0.5 and x=0, hence 3/2)x²+6x+6=(1/2)(3/2)v² and v=9.90ms/sec
(b) not sure what to do?

Thanks

By Brad Rodgers (P1930) on Saturday, November 25, 2000 - 10:36 pm :

I have a feeling I've made an error, but I'll go ahead and post just to see if I have.
here's my logic:
If

v=
dx/dt=3x² +6x+6

then

dt/dx=1/(3x² /2+6x+6

let u=3x² /2+6x+6

then dx=u^-1/2 du

So

t=òu^-1(u^-1/2 du)
Thus

t=-2u^-1/2 +c
=-2(3x² +6x+6)^-1/2 +c

At t=0, x=0

So

c=2/sqrt(6)

and at x=20,

this equals

t=2/(1326^1/2 +6^1/2 )

(remembering a sqrt can be either plus or minus)

This isn't a very nice solution, which leads me to believe I've made a mistake, but I can find none, so I'll post to see what others have to say about this. Sorry if this isn't much of a help, at least this method should work.

Brad

By Anonymous on Saturday, November 25, 2000 - 11:48 pm :

Hi Brad,

Thanks! I am gonna read through your logic to see if I understand. The answer that they give in the book is 1.70sec if thats any help.

By Susan Langley (Sml30) on Sunday, November 26, 2000 - 10:38 am :

Okay, I think I've come up with a niceish way of doing b):

a=F/m=(3x+6)/1.5=2x+4

so d² x/dt²-2x=4

Solve this differential equation:

Particular Integral is -2

Complimentary function is A e^{2^0.5t}+B e^{-2^0.5t}

So x=A e^{2^0.5t}+B e^{-2^0.5t}-2

But when t=0, x=0 and v=2^1.5

so A+B=2 and v=A×2^0.5e^{2^0.5t}-B×2^0.5 e^{-2^0.5t}

so 2^1.5=A×2^0.5-B×2^0.5 so A-B=2

so x=2e^{2^0.5t}-2

so at x=20, e^{2^0.5t}=11 so t=ln(11)/2^0.5=1.70 (3sf)

Hope this helps. Using this method does assume you have done some differential equations though.

By Susan Langley (Sml30) on Sunday, November 26, 2000 - 11:18 am :

Okay, I've now looked over what you two have done and have got some comments.

Firstly, anon: Why is c=6? Why can you differentiate over x less than 0? The initial conditions don't seem to allow that.

So this is I would integrate the energy change since time 0:

ò₀^x 3x+6 dx=0.5m v² -0.5m u² = (3/4)v² -(3/4)×8 (as u=2^1.5 so u² = 8)

so (3/2)x² +6x=(3/4)v² -6

v² = 2x² +8x+8

so when x=5 v² = 98 and v > 0 as accelaration always positive and initial v positive. so v=9.90(3sf). It looks like you just forgot about the plus, but this is a sounder method as you don't integrate below x=0. It's not that critical though...

So, Brad, where did that expression for v come from? Where did that expression for dx come from? There are also several arithmetical errors along the way. For instance in the final expression of t, you seem to have lost a minus sign (and no you shouldn't really just say the sqrt is plus or minus like that, you could have choosen the correct root earlier or mained a ± throughout) and the fact that they are two separate fractions.

Sorry to be harsh, so points are being fairly picky, so feel free to ignore those that seem unimportant if you want.

Susan

By Brad Rodgers (P1930) on Sunday, November 26, 2000 - 02:50 pm :

I got the expression for v from looking at anon's post and not seeing the 1/2 x 3/2v² part. Oops. (I was in somewhat of a hurry, so I didn't bother to do the integration myself) And I really hadn't had a problem like this before, so I had no idea what to do with the square roots.

Thanks for clearing that up,

Brad

By Anonymous on Sunday, November 26, 2000 - 03:23 pm :

Hi Susan,

c=6 came from the eqn when v^2 had still a coefficient to it. but the way you did this part (a) is also correct! since you do not get a constant of integration on int (F(x)) = 0.5mv² - 0.5mu² , no c exits!

I still don't understand how you solved part b), can anyone explain it any simpler methods. thanks,

By Kerwin Hui (Kwkh2) on Sunday, November 26, 2000 - 03:34 pm :

OK, we have

v(dv/dx)=4+2x, v=2Ö2 when x=0, giving

v² = 2(x+2)²

Square rooting both side and remember v is positive when x=0, we obtain

v=(x+2)×Ö2

but v=dx/dt, substitute and rearrange the equation to give

ò₀²⁰ dx/(x+2)=Ö2t

Evaluating the integral, we have

t= ___
Ö0.5

×ln(11)=1.70

seconds, to 3 s.f.

Kerwin

By Anonymous on Sunday, November 26, 2000 - 07:27 pm :

Gee whiz. Thanks, you people are great!
Thanks Kerwin!
Also thanks to Brad & Susan!

I think I sort of got into trouble when I thought "how do I get V on its own, so I can sub in dx/dt" I sqrooted the whole lot and saw it looked very messy. You factorised it good, i did'nt see that, becuase I left in in the form where v still had its coefficient. But now its all clear! Kerwin, in the step where you integrated, I am not sure why we don't include a constant on integration, can you briefly explain for me? Thanks.

By Kerwin Hui (Kwkh2) on Monday, November 27, 2000 - 01:01 pm :

Well, I am a bit lazy that I don't want to work out constants. Instead, I substitute the initial conditions as the lower limit for the integral, to save writing arbitrary constants all over the place.

Kerwin