Question

The displacement of a particle moving in a straight line is given by 1/t^2, where t is measured in seconds. Find the velocity of the particle at time t=a, t=1, t=2 and t=3

Answer 1

Do you know calculus?

Answer 2

A little. The semester just started.

Answer 3

\[v=\frac{ds}{dt}\] Where s is velocity, and s is displacement.

Answer 4

What's that odd notation? It means 'the change in s divided by the change in time' AT A PARTICULAR POINT (! the speed is not going to be uniform here!).

Answer 5

I'll walk you through it. \[s=\frac{1}{t^2}\]

Answer 6

Do you want the answer or to know a little more about how you get the answer/calculus?

Answer 7

I'd like to know the process. I understand the formula \[\frac{ f(a+h)-f(a) }{ h }\]

Answer 8

I understand it to a certain extent anyway.

Answer 9

All right, well, I'll continue @henpen 's response:
\[
\frac{d}{dx}f(x)=\lim_{\Delta x \to 0}\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)
\]Differentiation has a set of rules that make life easier, so I recommend you look those up, but, nevertheless, we continue. So, velocity is the change of displacement at a given instant of time, this could be interpreted to mean that velocity is the derivative of displacement, so, we say, for some velocity \(v\):
\[
v=\frac{d}{dt}\left(\frac{1}{t^2}\right)=
\frac{d}{dt}t^{-2}=-2t^{-3}=-\frac{2}{t^3}
\]Which is our velocity.
Now, we just plug in all of the values we need to find the answers to your questions.

Answer 10

I do not understand where your last set of equations came from. \[\frac{ d }{ dt }\]
How does this correlate to \[\frac{ 1 }{ t ^{2} }\] in order to be multiplied together?

Answer 11

It's not quite a multiplication- d/dt is the command : 'find the change of this thing as time goes on', roughly.

Answer 12

It's not a multiplication, you have to be careful. Like multiplication and addition, \(\frac{d}{dx}\) is an operator, which has two different arguments, a variable to differentiate over and a function which to differentiate.

Answer 13

Where did t^-2 become -2^3?

Answer 14

Why is it -2/t^3? I will explain (possibly concurrently with Lolwolf)

What is the change in something? On a graph, it is the gradient. In y=mx+c:
\[m=rise/run=dy/dx\]

Answer 15

Work it out using the limit, if you wish, I think it'd be an easier way to go. So, it would go as follows:
\[
\frac{d}{dt}\frac{1}{t^2}=\lim_{\Delta x \to 0}\left(\frac{(x+\Delta x)^{-2}-x^{-2}}{\Delta x}\right)=-2t^{-3}
\]

Answer 16

But, yes, imagine @henpen 's response, as the \(dx\) becomes infinitely small (That would be the \(\Delta x\) in the limit), and that would be your velocity *at a point*

Answer 17

What you really have to remember, though is that, like \(\Delta x\) the other part \(dx\) is understood to mean the same thing, just \(\textit{really}\) tiny.

Answer 18

But what about x^2 or something where the gradient (tangent to the curve) is always changing? We can't use a large triangle, too inaccurate: So make the triangle really small, so that dx and dy are almost negligible.