A very tricky problem that im just not getting

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In summary, the conversation is about a physics problem involving a glider on an inclined air track. The problem gives the initial and final velocity and asks for the angle of inclination. The individual has attempted to use the constant acceleration equation to solve for the angle, and is seeking help in finding a way to relate the angle to the acceleration or time. Another participant suggests that the acceleration is equal to g times the sine of the angle of inclination.
  • #1
ajc9387
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hey all

i have this very annoying physics problem that I've 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?

Someone please help!
 
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  • #2
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 there's a way to relate the sin of the angle to the acceleration or the time?
 
  • #3
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
 

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