# Motion on a parabola

1. Dec 3, 2009

### Drill

Hi everybody

we know that if we have an object sliding on a frictionless ramp,

the acceleration force will be constant, and it equals to

a=g * sin(theta)
where theta is the ramp angle w.r.t. the ground

so the path of motion in this problem can be written mathematically as a linear function
y(x)=bx
and hence the tangent tan(theta)= bx /x =b

The Question is
if the path of motion is parabolic and is of the form

y(x) = a x^2

how to solve for the acceleration with respect to time ???????

be aware that ,in this case the acceleration is not constant , and it always equals to the tangent of the parabola at the current location of the object.
and the tangent in this case is the first derivative of y which is
y'=2ax

as we see the tangent and hence the acceleration is a function of x

so ,

how to calculate the time as the function of position ??

thanks

Last edited: Dec 3, 2009
2. Dec 3, 2009

### rock.freak667

Using n-t coordinates you'd get that the acceleration vector is

$$\vec{a}=\dot{v} \hat{e_t}+ \frac{v^2}{\rho} \hat{e_n}$$

where

$$\rho =\frac{(1+(y')^2)^\frac{3}{2}}{y''}$$

Not sure if I wrote down the formula for ρ correctly though.

then again, for a time parameter x=t.

3. Dec 4, 2009

### Drill

Hi

and thanks for replying

but can you explaine how this equation came about

$$\vec{a}=\dot{v} \hat{e_t}+ \frac{v^2}{\rho} \hat{e_n}$$

or at least show me the source link

4. Dec 4, 2009