Physics Problem: Displacement, Acceleration, Velocity

[shawonna23]

A skateboarder, starting from rest, rolls down a 11.0 m ramp. When she arrives at the bottom of the ramp her speed is 7.20 m/s.

(a) Determine the magnitude of her acceleration, assumed to be constant.

(b) If the ramp is inclined at 24.5° with respect to the ground, what is the component of her acceleration that is parallel to the ground?

 

[bullet_ballet]

(a) Since the skateboarder is on an incline, a = g sinθ. To find the time it takes to descend, find the average velocity v = (vf - vi)/2. We know that t = distance/velocity, or in this case, 11/3.6. Using a = v/t, you can find that θ = ArcSin v/gt.

(b) If you draw a diagram of the problem you'll see that the x-component of acceleration is ax = a cos24.5

 

[sher68pk]

Average Velocity = (vf+vi)/2 and acceleration = (vf-vi)/t please could you explain how average velocity has been expressed as vf-vi over 2

 

[sher68pk]

Displacement Problem:
A car travels at a speed of 5 m/s along a square root ABCD find its displacement and average velocity from A to B, from A to C and from A to D i.e from A to all the corners with length of one side of the square is = 20 m.
Find dispacement and average velocity:
i- from A to B
ii- form A to C
iii- from A to D

 

[asteriatic]

(a) To determine the magnitude of the skateboarder's acceleration, we can use the equation a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity (in this case, 0 m/s), and t is the time it takes for the skateboarder to travel the 11.0 m ramp. Since we are given the final velocity of 7.20 m/s and the distance of 11.0 m, we can rearrange the equation to solve for acceleration: a = (7.20 m/s - 0 m/s)/t. This simplifies to a = 7.20 m/s^2. Therefore, the magnitude of the skateboarder's acceleration is 7.20 m/s^2.

(b) To find the component of the skateboarder's acceleration that is parallel to the ground, we can use the equation a = gsinθ, where g is the acceleration due to gravity (9.8 m/s^2) and θ is the angle of the ramp (24.5°). Plugging in these values, we get a = (9.8 m/s^2)(sin 24.5°) = 4.20 m/s^2. This means that 4.20 m/s^2 of the skateboarder's acceleration is parallel to the ground.