Calculating Acceleration of Skateboarder on 11.9m Ramp

[miniz88]

Homework Statement

A skateboarder, starting from rest, rolls down a 11.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.77 m/s. (a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 29.8 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground?

The Attempt at a Solution

I don't even have a clue where to start? How can i figure out the magnitude of acceleration without the time?? And i dnt even understand what the second part is asking at all?

Any help on where to start is appreciated..

 

[hage567]

I don't even have a clue where to start? How can i figure out the magnitude of acceleration without the time??

You may not have the time, but you have the distance. Take a closer look at your kinematic equations.

And i dnt even understand what the second part is asking at all?

Since she is accelerating down the ramp (so at an angle with the ground). You can resolve her acceleration into horizontal and vertical components.

 

[Moetasim]

I would suggest starting by breaking down the problem into smaller parts and using the equations of motion to solve for the unknowns.

For part (a), we know that the skateboarder starts from rest and reaches a speed of 6.77 m/s at the bottom of the ramp. We also know the distance traveled, which is 11.9 m. The equation we can use to solve for acceleration is v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (in this case, 0 m/s), a is the acceleration, and s is the distance traveled. Rearranging the equation, we get a = (v^2 - u^2)/2s. Plugging in the values, we get a = (6.77^2 - 0^2)/2(11.9) = 2.88 m/s^2. This is the magnitude of the acceleration.

For part (b), we need to find the component of the acceleration that is parallel to the ground. This means we only need to consider the acceleration in the x-direction (since the ramp is inclined at an angle with respect 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 (29.8°). Plugging in the values, we get a = (9.8)(sin 29.8°) = 4.99 m/s^2. This is the component of acceleration parallel to the ground.

Overall, it is important to remember to carefully read and understand the problem, break it down into smaller parts, and use relevant equations to solve for the unknowns.