# Δt and Limit of Δx/Δt as Δt -> 0 at t=3?

1. Mar 22, 2013

### ACSC

1. The problem statement, all variables and given/known data
The position of a particle moving along the x axis varies in time according to the expression x = 5t^2, where x is in meters and t is in seconds. Evaluate its position at the following times.
(a) t = 3.00 s
x = ? m
(b) t = 3.00 s + Δt
xf(final x) = ? m
(c) Evaluate the limit of Δx/Δt as Δt approaches zero to find the velocity at t = 3.00 s.
v = ? m/s

2. Relevant equations
Maybe lim x->0 ∫Δx/Δt

3. The attempt at a solution
(a) x = 45m, got it.
(b) I don't understand what it's asking me.
(c) lim t-> 0 ∫ 5x^2 Δx/Δt at t=3
= lim t-> 0 [10t]
I don't know what to do from here since it asks me to find velocity when t->0 at when t=3 at the same time.

2. Mar 22, 2013

### BruceW

for part b, you are given t = 3 + Δt and the equation x=5t^2 So just use this. (Don't worry that your final answer depends on Δt - that is what they want).

p.s. Welcome to physicsforums !

3. Mar 22, 2013

### ACSC

I'm still not quite sure what to do. Do I make the t in the x equation the subject the simultaneously solve it? Isn't (b) asking for the distance travelled rather than Δt?

This feels like a really simple question that I'm supposed to understand straightaway. I've done basic physics and that doesn't seem to help...

Thank you for the welcome. :)

(It's 12AM here so I probably won't be able to respond until my tomorrow morning.)

4. Mar 22, 2013

### haruspex

b) is asking you to substitute 300+Δt for t and xf for x in the expression x = 5t2.
There's a step they've left out in going from b) to c). You need to substitute x+Δx for xf. (I don't know why they said xf instead of x+Δx in the first place.)
No, it's as Δt→0. Δt is not Δ multiplied by t, it's a separate variable representing a small change in t.

5. Mar 22, 2013

### ACSC

I think that makes sense, thank you.

Would (b) be 144m?
Is the second working out for (c) correct?
Working out picture here.

Last edited: Mar 22, 2013
6. Mar 22, 2013

### haruspex

No, it should be an algebraic expression involving the unknown Δt. It's the fourth line of the working you attached (except there's a mistake in the last term).
In part c, x is the same as in part a (=45m). We are concerned with what happens over the time interval from 3 seconds to 3 +Δt seconds. At 3 seconds, x = 45m. At 3+Δt seconds it's x+Δx, the answer to part b. To get the change in x, you subtract the x to leave just the Δx. So you get Δx = 30Δt + 5Δt2.
For c, it doesn't ask for the limit of Δx as Δt→0, it asks for the limit of Δx/Δt. So divide through by Δt before taking the limit.

7. Mar 22, 2013

### ACSC

Oh, I forgot to times the 5 into the last part. Thanks.
xf = 5Δt2 + 30Δt + 45 ?
The question wants the answer in meters though. :S

If I sub in Δt=3, I would get 180m. Would that be it? Seems like there's something wrong in subbing Δt=3, though.

I think I get what you mean.
Like this?

8. Mar 22, 2013

### haruspex

Strange. That isn't possible unless they give you a value to use for Δt.
That's it.

9. Mar 22, 2013

### ACSC

Can we assumed that ti=0 and tf=3? Though that would be the exact same thing as t=3, wouldn't it?
I'll email my teacher and ask what they want the form of the answer to be.

GREAT! Thank you so much for guiding me. :)

10. Mar 24, 2013

### BruceW

No, their notation is pretty bad, to be honest. They start in part a) with:
t = 3.00 s
Which is fine so far. But then they use:
t = 3.00 s + Δt
They are using the same variable name for two different things. What they should have written is something like:
ti = 3.00 s
tf = 3.00 s + Δt
This is what they really mean.

edit: (and by this notation, the particle is at x at time ti and it is at xf at time tf)
edit again: and so, by this, you can probably see that in part b) it is not possible to give a numerical answer because Δt is an unknown variable. Also, don't let the notation Δt put you off. It is just like any normal variable. You can call it 'a' or whatever, it does not have any special properties. (I think haruspex already said this, but it is a good point).

Last edited: Mar 24, 2013