i don't understand how to do this question, its a phys 11:
"A skier is acc. down a 30.0 degree hill at 3.80m/s^2"
a) What is the horiz. comp. of the acc.?
b) What is the vert. comp. of the acc.?
c) Assuming she starts from rest and acc. uniformly down hill, how long will it take for her to reach bottom if the elev. change is 355m?
a) 3.80 cos 30
b) -3.80 sin 30 if + is up
c) only vertical component matters for change in elevation
y = Yo + Vo t + (1/2) a t^2
-355 = 0 + 0 t + (.5)(-1.90) t^2
you can solve that I am sure.
thanks
To solve this problem, you'll need to use trigonometry and the kinematic equations for linear motion. Let's break down each part of the question:
a) To find the horizontal component of acceleration, we need to consider the angle of the hill. The horizontal direction is perpendicular to the slope. Since the skier is accelerating down the hill, the horizontal component of acceleration is zero. This is because all of the acceleration is in the vertical direction.
b) To find the vertical component of acceleration, we can use trigonometry. Given that the hill is inclined at an angle of 30.0 degrees, we can use the equation:
Vertical acceleration (A_y) = Acceleration (A) * sin(angle)
A_y = 3.80 m/s^2 * sin(30.0 degrees)
Now you can calculate the vertical component of acceleration.
c) To find the time it takes for the skier to reach the bottom of the hill, we can use the kinematic equation:
Distance (d) = Initial velocity (v_i) * time (t) + (1/2) * acceleration (a) * t^2
In this case, the initial velocity (v_i) is zero since the skier starts from rest. The distance is given as 355 meters, and the acceleration (a) is the total acceleration down the hill.
355 m = 0 * t + (1/2) * 3.80 m/s^2 * t^2
Now we can solve for time (t) using this equation and find out how long it will take for the skier to reach the bottom of the hill.
Remember to label your variables clearly, use the appropriate units, and double-check your calculations to ensure accuracy.