Displacement homework help

1. Aug 30, 2006

konichiwa2x

The displacement 'x' and time 't' of a particle are related as follows:

t = $$\alpha$$$$x^2$$ + $$\beta$$$$x$$
where alpha and beta are constants
Find the retardation of the body in terms of 'v'
Can someone tell me how to do this??

Last edited: Aug 30, 2006
2. Aug 30, 2006

J77

If v is the velocity, you may want to look at differentiating...

3. Aug 30, 2006

Astronuc

Staff Emeritus
Are the two x's the same dimension, or do they necessarily have the same exponent?

Otherwise $\alpha x^2\,+\,\beta x^2$ would simply to

$(\alpha\,+\,\beta) x^2$

4. Aug 30, 2006

konichiwa2x

sorry there was not meant to be an exponent for the second 'x'. I have tried differentiating, but keep getting the wrong answer. I got acc = -2(alpha)v^2/[2(alpha)x + beta]is it right?

Last edited: Aug 30, 2006
5. Aug 30, 2006

Astronuc

Staff Emeritus
So just to be clear, t = $\alpha x^2\,+\,\beta x$?

So differentiating as suggested by J77, would yield

1 = $\alpha\,(2x)\,\dot{x}\,+\,\beta$

Then separate to find v = dx/dt

If it is $$\beta^x$$, i.e. ß^x, that is somewhat more complicated.

Last edited: Aug 30, 2006
6. Aug 30, 2006

konichiwa2x

Sorry I dont get it. and what do the dot above the 'x' indicate?? And it is $$\beta$$x

7. Aug 30, 2006

Astronuc

Staff Emeritus
$$\dot{x}$$ = dx/dt = v

What do you know about retardation? Do you have a definition or expression for it?

8. Aug 31, 2006

konichiwa2x

retardation is just negative acceleration right?
anyway I have progressed. can you check if this is correct?
a is the acceleration.

t = $$\alpha x^2+ \beta x$$
1 = $$2\alpha xv+ \beta v$$
0 = $$2\alpha(xa + v^2)+\beta a$$

therfore, a = $$\frac{-2v^2}{2 \alpha x + b}$$

Last edited: Aug 31, 2006
9. Aug 31, 2006

J77

Looks good - but you forgot an alpha on the top

(and your beta seems to have become a b)