Physics Problem: Displacement, Acceleration, Velocity

In summary, the skateboarder starts from rest and rolls down a 11.0 m ramp, reaching a speed of 7.20 m/s at the bottom. The magnitude of her acceleration, assumed constant, can be determined using a = v/t. The component of her acceleration parallel to the ground can be found using a = g sinθ. To find the time it takes to descend, the average velocity is calculated as (vf + vi)/2 and used in the equation θ = ArcSin v/gt. To find displacement and average velocity of a car traveling at 5 m/s along a square root ABCD, with a side length of 20 m: - From A to B: The displacement is 20
  • #1
shawonna23
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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?
 
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  • #2
(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
 
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  • #3
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
 
  • #4
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
 
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  • #5


(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.
 

1. What is displacement in physics?

Displacement in physics refers to the distance and direction of an object's change in position from its starting point.

2. How is acceleration calculated in physics?

Acceleration in physics is calculated by dividing the change in velocity by the time it took for that change to occur. The formula for acceleration is a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

3. What is the difference between speed and velocity?

In physics, speed refers to the rate at which an object is moving, while velocity refers to the rate at which an object is changing its position. Velocity takes into account the direction of the object's movement, while speed does not.

4. How does displacement differ from distance?

Displacement and distance are both measures of an object's position, but they differ in that displacement takes into account the direction of an object's change in position, while distance only considers the total length of the path traveled by the object.

5. What is the relationship between displacement, velocity, and acceleration?

Displacement, velocity, and acceleration are all related in that acceleration is the rate of change of velocity over time, and velocity is the rate of change of displacement over time. In other words, acceleration is the second derivative of displacement with respect to time.

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