# 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

### Staff: Mentor

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

### Staff: Mentor

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

### Staff: Mentor

$$\dot{x}$$ = 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)