A skier starts from rest, accelerates down 30degree slope

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In summary, the skier starts from rest at the top of a hill of height h and accelerates down a 30.0 degree ski slope with an acceleration of a = 4.9 m/s^2. Using the suvat equations, the skier's instantaneous velocity at the bottom of the hill is 63.25 m/s and her average velocity down the slope is 31.64 m/s. At the base of the hill, the skier continues horizontally for another 250 m before coming to a stop. Her total vector displacement from start to finish is 750 m.
  • #1
punyhuman92
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1. A skier starts from rest at the top of a hill of height h down a 30.0 degree ski slope with an acceleration a.

a.) What is her speed at the bottom of the hill?
b.) If the hill was h = 250. m high and her acceleration down the slope was a = 4.00 m/s2, what was instantaneous speed at the bottom of the hill, and her average speed down the slope?
c.) At the base of the hill, she continues horizontally for another 250 m before coming to a stop and ending her run. What is her total vector displacement from start to finish?

I'm asking about a) for now.

for a.) I drew the FBD, rotated it so Fgx is parallel to the slope of the hill.
mgSin30=Fgx
Fgx=4.9m

then I did Fnet/m=a
the m's cancel so I'm left with a=4.9 m/s^2
it's asking for the speed though, aka velocity so is my answer expected to simply be 4.9m/s^2*t? I feel like I should be able to work with it more but i don't see how I could?
 
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  • #2
For part b) I used the formula Vf^2-Vi^2=2a(\delta)x

For the change in x I found the hypotenuse of the triangle, is that correct? or should I have used the 250m that was given?

After plugging in the values I got 63.25 m/s for my instantaneous velocity.
 
  • #3
I'm not sure how to find the average velocity. Unless all I have to do is V=at? In which case I get 31.64 m/s
 
  • #4
I know its early but can anyone else verify if the work I'm doing is correct? :(
 
  • #5
For C) For letter C is it simply 750m?
 
  • #6
Acceleration isn't velocity, you have to work out velocity. I think you need to use the suvats.
s=displacement(m)
u=start velocity(ms-1)
v=end velocity(ms-1)
a=acceleration(ms-2)
t=time(s)
There are 5 suvat equations, each omits one of the 5 measurements. Depending on the information given, you use a different suvat equation.
v=u+at (no s)
s=vt-0.5at2 (no u)
s=ut+0.5at2 (no v)
s=0.5t(u+v) (no a)
v2=u2+2as (no t)
Remember:These equations can only be used if acceleration is constant.
Where it stops (e.g. the point turns around mid-air, or via the force of friction) then v=0.
If it starts from rest, u=0.
s is displacement, not velocity.
Any one of the measurements can be negative or come out negative.
If s or t comes as two answers, this is because it passes the same point twice.

I gave you more than the information you need to solve this problem, but suvats will come in useful in all sorts of places, so learning how to use them is a good idea.
 
  • #7
jetwaterluffy said:
Acceleration isn't velocity, you have to work out velocity. I think you need to use the suvats.
s=displacement(m)
u=start velocity(ms-1)
v=end velocity(ms-1)
a=acceleration(ms-2)
t=time(s)
There are 5 suvat equations, each omits one of the 5 measurements. Depending on the information given, you use a different suvat equation.
v=u+at (no s)
s=vt-0.5at2 (no u)
s=ut+0.5at2 (no v)
s=0.5t(u+v) (no a)
v2=u2+2as (no t)
Remember:These equations can only be used if acceleration is constant.
Where it stops (e.g. the point turns around mid-air, or via the force of friction) then v=0.
If it starts from rest, u=0.
s is displacement, not velocity.
Any one of the measurements can be negative or come out negative.
If s or t comes as two answers, this is because it passes the same point twice.

I gave you more than the information you need to solve this problem, but suvats will come in useful in all sorts of places, so learning how to use them is a good idea.

I understand that, but velocity is equal to acceleration multiplied by time, but how do I find time? I feel like I am not given enough to solve the problem, i ended up with (t)(4.9m/s^2) as my velocity

I don't have the displacement or distance, neither do i have the time. so how can i find the velocity with such little information, unless the question itself doesn't expect me to do anything more than simplify?
 
  • #8
punyhuman92 said:
I don't have the displacement or distance,
Yes you do.
punyhuman92 said:
If the hill was h = 250. m high
Even if you didn't, you can still give your answer in terms of height or time.
 
  • #9
I thought a) and b) and c) where all different questions pertaining to the same problem and unrelated, fml you might be onto something.
 
  • #10
punyhuman92 said:
I thought a) and b) and c) where all different questions pertaining to the same problem and unrelated, fml you might be onto something.

Sometimes they make mistakes with questions and their orders. Is this an offical exam paper or just a question given by your book or teacher?
 
  • #11
A question given by my professor, I have an exam in two hours and i procrastinated looking at his review sheet. i emailed him but i doubt he'll get back to me before the time of the test. (i've been studying our past quizzes/hw instead for the past week)
 
  • #12
I still think each question is separate though i wish the question was more specific.
 
  • #13
Question: those suvat equations, are those used to find the average or instantaneous?

What would I use to find inst. velocity vs avg velocity?
 
  • #14
punyhuman92 said:
Question: those suvat equations, are those used to find the average or instantaneous?

What would I use to find inst. velocity vs avg velocity?
Instantaneous. Average velocity is the midpoint of u and v.
 

1. How does the angle of the slope affect the skier's acceleration?

The angle of the slope affects the skier's acceleration by determining the magnitude and direction of the gravitational force acting on the skier. A steeper slope will result in a larger component of the gravitational force acting parallel to the slope, causing the skier to accelerate faster.

2. What factors besides the slope angle can affect the skier's acceleration?

Besides the slope angle, the skier's mass, air resistance, and friction with the snow can also affect their acceleration. A heavier skier will experience a greater gravitational force, resulting in a larger acceleration. Air resistance can also slow down the skier's acceleration, while friction with the snow can either increase or decrease their acceleration depending on the texture of the snow and the skier's technique.

3. How does the skier's starting speed affect their acceleration down the slope?

The skier's starting speed does not directly affect their acceleration down the slope, as their acceleration will be determined by the slope angle and other factors. However, a higher starting speed may cause the skier to reach a higher maximum velocity, as they will have more kinetic energy to convert into speed while accelerating down the slope.

4. Is the skier's acceleration constant or does it change as they go down the slope?

The skier's acceleration is not constant, as it will change depending on the slope angle and other factors. As the skier goes down the slope and their speed increases, the magnitude of their acceleration will decrease due to air resistance and friction with the snow. However, if the slope angle changes, their acceleration will also change to reflect the new gravitational force acting on the skier.

5. How can the skier's acceleration be calculated on a 30 degree slope?

The skier's acceleration can be calculated using the formula a = gsin(θ), where a is the acceleration, g is the gravitational acceleration (9.8 m/s^2), and θ is the slope angle in degrees. In this case, the acceleration on a 30 degree slope would be approximately 4.9 m/s^2.

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