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 = \alphax^2 + \betax
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 = \alphax^2 + \betax^2

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

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

 

[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 = \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.

 

[konichiwa2x]

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

 

[Astronuc]

\dot{x} = 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 = \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}

 

[J77]

Looks good - but you forgot an alpha on the top

(and your beta seems to have become a b)