(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 8.86 m/s when going down a slope for 3.30 s? (b) How far does the skier travel in this time?

a. a = (Vf - Vo) / t,

a = (8.86 _ 0) / 3.3 = 2.68m/s^2.

b. d = Vo + 0.5a*t^2,
d = 0 + 1.34*(3.3)^2 = 14.62m.

To find the magnitude of the average acceleration of the skier, we can use the formula for average acceleration:

Average acceleration = (final velocity - initial velocity) / time

In this case, the skier starts from rest, so the initial velocity (u) is 0 m/s. The final velocity (v) is 8.86 m/s, and the time (t) is 3.30 s.

(a) To find the magnitude of the average acceleration:

Average acceleration = (8.86 m/s - 0 m/s) / 3.30 s = 2.68 m/s^2

So, the magnitude of the average acceleration of the skier is 2.68 m/s^2.

To find how far the skier travels in this time, we can use the equation for distance traveled:

Distance = (initial velocity + final velocity) / 2 * time

In this case, the initial velocity (u) is 0 m/s, the final velocity (v) is 8.86 m/s, and the time (t) is 3.30 s.

(b) To find the distance traveled:

Distance = (0 m/s + 8.86 m/s) / 2 * 3.30 s = 14.58 m

So, the skier travels a distance of 14.58 meters in 3.30 seconds.