- #1

- 16

- 0

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?

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?