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.