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(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 10.8 m/s when going down a slope for 3.36 s? (b) How far does the skier travel in this time?

(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 10.8 m/s when going down a slope for 3.36 s? (b) How far does the skier travel in this time?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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(a) What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 10.8 m/s when going down a slope for 3.36 s? (b) How far does the skier travel in this time?

### Problem Statement: (a) **What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 10.8 m/s when going down a slope for 3.36 seconds?** (b) **How far does the skier travel in this time?** ### Solution Fields: (a) Number: Units: (b) Number: Units: ### Explanation: To solve this problem: #### Part (a): 1. **Average Acceleration (a):** - **Formula:** \( a = \frac{v_f - v_i}{t} \) - \( v_f \) is the final velocity (10.8 m/s) - \( v_i \) is the initial velocity (0 m/s, since the skier starts from rest) - \( t \) is the time (3.36 s) 2. **Calculation:** - \( a = \frac{10.8 \, \text{m/s} - 0 \, \text{m/s}}{3.36 \, \text{s}} \) #### Part (b): 1. **Distance Traveled (d):** - **Formula:** \( d = v_i t + \frac{1}{2} a t^2 \) - Again, \( v_i \) = 0 m/s - \( t \) = 3.36 s - Use the acceleration \( a \) found in part (a) 2. **Calculation:** - \( d = 0 \cdot 3.36 + \frac{1}{2} \cdot a \cdot (3.36)^2 \)
Transcribed Image Text:### Problem Statement: (a) **What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of 10.8 m/s when going down a slope for 3.36 seconds?** (b) **How far does the skier travel in this time?** ### Solution Fields: (a) Number: Units: (b) Number: Units: ### Explanation: To solve this problem: #### Part (a): 1. **Average Acceleration (a):** - **Formula:** \( a = \frac{v_f - v_i}{t} \) - \( v_f \) is the final velocity (10.8 m/s) - \( v_i \) is the initial velocity (0 m/s, since the skier starts from rest) - \( t \) is the time (3.36 s) 2. **Calculation:** - \( a = \frac{10.8 \, \text{m/s} - 0 \, \text{m/s}}{3.36 \, \text{s}} \) #### Part (b): 1. **Distance Traveled (d):** - **Formula:** \( d = v_i t + \frac{1}{2} a t^2 \) - Again, \( v_i \) = 0 m/s - \( t \) = 3.36 s - Use the acceleration \( a \) found in part (a) 2. **Calculation:** - \( d = 0 \cdot 3.36 + \frac{1}{2} \cdot a \cdot (3.36)^2 \)
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