(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 8.02 m/s when going down a slope for 4.28 s? (b) How far does the skier travel in this time?
a=v/t
s=at²/2
To solve this problem, we need to use the kinematic equation that relates acceleration, initial velocity, time, and displacement:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
(a) To find the magnitude of the average acceleration, we can rearrange the equation and solve for acceleration:
a = (v - u) / t
Given:
u = 0 m/s (starting from rest)
v = 8.02 m/s
t = 4.28 s
Substituting the values into the equation:
a = (8.02 m/s - 0 m/s) / 4.28 s
a = 8.02 m/s / 4.28 s
a ≈ 1.87 m/s²
Therefore, the magnitude of the average acceleration of the skier is approximately 1.87 m/s².
(b) To find the distance traveled by the skier, we can use another kinematic equation:
s = ut + (1/2)at²
Since the skier starts from rest, the initial velocity u is 0.
s = (1/2)at²
Substituting the known values:
s = (1/2)(1.87 m/s²)(4.28 s)²
s = (1/2)(1.87 m/s²)(4.28 s)(4.28 s)
s ≈ 18.96 m
Therefore, the skier travels approximately 18.96 meters in this time.