What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 5.8 m/s when going down a slope for 2.8 s? How far does the skier travel in this time?

acceleration=changevelocity/time=5.8/2.8 m/s^2

d= 1/2 a t^2

13.645

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

acceleration = (final velocity - initial velocity) / time

Given:
Initial velocity (u) = 0 m/s (starting from rest)
Final velocity (v) = 5.8 m/s
Time (t) = 2.8 seconds

Substituting these values into the formula, we have:

acceleration = (5.8 m/s - 0 m/s) / 2.8 s
acceleration = 5.8 m/s / 2.8 s
acceleration ≈ 2.07 m/s²

Therefore, the magnitude of the average acceleration of the skier is approximately 2.07 m/s².

Now, to find the distance traveled by the skier, we can use the equation of motion:

distance = (initial velocity * time) + (0.5 * acceleration * time²)

Given:
Initial velocity (u) = 0 m/s
Acceleration (a) = 2.07 m/s²
Time (t) = 2.8 seconds

Substituting these values into the equation, we have:

distance = (0 m/s * 2.8 s) + (0.5 * 2.07 m/s² * (2.8 s)²)
distance = 0 m + (0.5 * 2.07 m/s² * 7.84 s²)
distance = 0 m + (0.5 * 2.07 m/s² * 61.4656 s²)
distance = 0 m + 63.7258 m
distance ≈ 63.73 m

Therefore, the skier travels approximately 63.73 meters in the given time.