# A very tricky problem that im just not getting

1. Feb 13, 2006

### ajc9387

hey all

i have this very annoying physics problem that ive been looking at for hours.

In the physics laboratory, a glider is released from rest on a frictionless air track inclined at an angle. If the glider has gained a speed of 22.0 cm/s in traveling 40.0 cm from the starting point, what was the angle of inclination of the track? Assume that the initial velocity is zero.

I understand that you have the initial velocity and the final velocity, and you can find the acceleration with a "constant acceleration" equation, specifically DeltaX= 1/2acceleration*time squared, but how on earth do you find the angle of inclination given only the value for x, and not for y or the hypotenuse?

2. Feb 13, 2006

### ajc9387

oh....my work

So through using said acceleration equation
(v squared = v initial squared plus 2(acceleration)(displacement)

so i got that the acceleration is 6.05 cm/s

then from that i got that the time is equal to 22= 6.05t

So t is 3.64 seconds.

So i'm guessing theres a way to relate the sin of the angle to the acceleration or the time?

3. Feb 13, 2006

### chroot

Staff Emeritus
If the angle of inclination were 90 degrees, the glider would be falling straight down, accelerating at g=9.8 m/s^2. If the angle of inclination were 0 degrees, the glider would be sitting still, with an acceleration of 0.

The acceleration is therefore equal to g times the sine of the angle of inclination.

- Warren