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A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity of the skier at the bottom of the hill



Answer :

eternal3fear
we can use the kinematic equation
[tex](v_f)^2 = (v_i)^2 + 2*a*d[/tex]
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]

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