(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 10.5 m/s when going down a slope for 3.18 s? (b) How far does the skier travel in this time?

a. a = 10.5/3.18 = 3.3 m/s^2.

b. d = 0.5a*t^2 = 0.5*3.3*3.18^2 = 16.7 m.

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

average acceleration (a) = (final velocity - initial velocity) / time

Let's break down the problem step by step:

(a) Magnitude of average acceleration:
1. The skier starts from rest, so the initial velocity (u) is 0 m/s.
2. The final velocity (v) is given as 10.5 m/s.
3. The time (t) is given as 3.18 s.
4. Plug the values into the formula:

a = (v - u) / t
= (10.5 m/s - 0 m/s) / 3.18 s
= 10.5 m/s / 3.18 s
≈ 3.30 m/s²

Thus, the magnitude of the average acceleration of the skier is approximately 3.30 m/s².

(b) Distance traveled by the skier:
To find the distance traveled by the skier in this time, we can use the formula:

distance (d) = initial velocity * time + (1/2) * average acceleration * time²

1. The initial velocity (u) is 0 m/s since the skier starts from rest.
2. The average acceleration (a) is 3.30 m/s², as calculated in part (a).
3. The time (t) is given as 3.18 s.
4. Plug the values into the formula:

d = 0 m/s * 3.18 s + (1/2) * 3.30 m/s² * (3.18 s)²
= 0 m + (1/2) * 3.30 m/s² * 10.1284 s²
= 0 m + 5.0642 m²/s²
≈ 5.06 m (rounded to two decimal places)

Hence, the skier travels approximately 5.06 meters in this time.