- #1

rmawatson

- 3

- 0

## Homework Statement

a particle on a slope with angle theta with no friction. v(0) = 0, x(0) = 0, with coordinates i down the slope and j normal to it.

I am confused about why with the constant velocity forumla I get a different answer to my attempted method.. I can't see what's wrong..I need to find the velocity at "l"

x is the top of the slope of a particle on a smooth surface, with no friction,

v(0) = 0, x(0) = 0

along i direction I am starting with:

mgsin(theta) == ma

gsin(theta) == a

using constant acceleration formula

v^2 = v0^2 + 2a0(x-x0)

v^2 = 0 + 2gsin(theta)(x-0)

v = sqrt( 2gl*sin(theta) )

My original attempt below is wrong, but I can't see why. I want to know what it doesn't work the same.

so from

gsin(theta) == a

Integrating wrt t

gsin(theta)t == v + c

Integrating wrt t again

1/2*gsin(theta)t^2 == x + ct + d

with v(0) == 0 and x(0) == 0

0 = c and d = 0

so if I now plugged in 'l' to the equation for position

1/2*gsin(theta)t^2 == l

and solve for t I get,

t = sqrt[ (2l)/(gsin(theta)) ]

so this is the time at which position == l ?

If I then plug this time into the equation for velocity,

gsin(theta)t == v

gsin(theta)*sqrt[ (2l)/(gsin(theta)) ] = v

not the same as with the constant velocity forumla..

why ? what is wrong with this method

Thanks for any help