# A skier starts from rest, accelerates down 30degree slope

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?

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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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.

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

I know its early but can anyone else verify if the work I'm doing is correct? :(

For C) For letter C is it simply 750m?

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.

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 im 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?

I don't have the displacement or distance,
Yes you do.
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.

I thought a) and b) and c) where all different questions pertaining to the same problem and unrelated, fml you might be onto something.

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?

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)

I still think each question is separate though i wish the question was more specific.

Question: those suvat equations, are those used to find the average or instantaneous?

What would I use to find inst. velocity vs avg velocity?

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.