What is the Relationship Between Displacement and Time in Particle Motion?

[konichiwa2x]

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

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

 

[J77]

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

t = [tex]\alpha[/tex][tex]x^2[/tex] + [tex]\beta[/tex][tex]x^2[/tex]

Find the retardation of the body in terms of 'v'
Can someone tell me how to do this??

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

 

[Astronuc]

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

Otherwise [itex]\alpha x^2\,+\,\beta x^2[/itex] would simply to

[itex](\alpha\,+\,\beta) x^2[/itex]

 

[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?

 

[Astronuc]

So just to be clear, t = [itex]\alpha x^2\,+\,\beta x[/itex]?

So differentiating as suggested by J77, would yield

1 = [itex]\alpha\,(2x)\,\dot{x}\,+\,\beta[/itex]

Then separate to find v = dx/dt

If it is [tex]\beta^x[/tex], i.e. ß^x, that is somewhat more complicated.

 

[konichiwa2x]

Sorry I don't get it. and what do the dot above the 'x' indicate?? And it is [tex]\beta[/tex]x
can u please explain?

 

[Astronuc]

[tex]\dot{x}[/tex] = dx/dt = v

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

 

[konichiwa2x]

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

t = [tex]\alpha x^2+ \beta x[/tex]
1 = [tex]2\alpha xv+ \beta v[/tex]
0 = [tex]2\alpha(xa + v^2)+\beta a[/tex]

therfore, a = [tex]\frac{-2v^2}{2 \alpha x + b} [/tex]

 

[J77]

Looks good - but you forgot an alpha on the top

(and your beta seems to have become a b)