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ame: Page 34- Practice Problem 2.2: What is the cheetah's average speed over (a) the first second of the attack and (b) the first two seconds of the attack? Answers: (a) 5.00m/s, (b) 10.0m/s.

ame: Page 34- Practice Problem 2.2: What is the cheetah's average speed over (a) the first second of the attack and (b) the first two seconds of the attack? Answers: (a) 5.00m/s, (b) 10.0m/s.

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) The situation (b) Our sketch Vehicle ot Ax Uav,x= = At 2 We choose to place the origin at the vehicle. At time 12 = 2.00 s, the cheetah's position x2 is x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m. The displacement during this interval is SOLUTION SET UP Figure 2.10b shows the diagram we sketch. First we define a coordinate system, orienting it so that the cheetah runs in the +x direc- tion. We add the points we are interested in, the values we know, and the values we will need to find for parts (a) and (b). 40.0 m - 25.0 m 2.00 s-1.00 s We draw an axis. We point it in the direction the cheetah runs, so that our values will be positive. A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw. Cheetah Cheetah starts = 4-0 xo-20.0 m SOLVE Part (a): To find the displacement Ax, we first find the so that cheetah's positions (the values of x) at time t₁ = 1.00 s and at time 12=2.00 s by substituting the values of t into the given equation. At time = 1.00 s, the cheetah's position x₁ is x₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m. 0.02 15.0 m. → x₁ = ? 1₁-1.00 s We mark the initial positions of the cheetah and the antelope. (We won't use the antelope's position-but we don't know that yet.) Ax= x₂ = x₁ = 40.0 m - 25.0 m Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can now find the average velocity for that interval: 15.0 m = 15.0 m/s. 1.00 s Ax-? Vov.x => 807 X₂=? 1₂-2.00 s We're interested in the cheetah's motion between 1s and 2 s after it begins running. We place dots to represent those points. Antelope Uav,.x = Ax At Antelope 50.0 m x (m) x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m, 26.05 m 1.10 s 5 We add symbols for known and unknown quantities. We use subscripts 1 and 2 for the points at t=1s and f= 2 s. Part (c): The instantaneous velocity at 1.00 s is approximately (but exactly) equal to the average velocity in the interval from t₁ =- t₂ = 1.10 s (i.e., At = 0.10 s). At 1₂ = 1.10 s, 25.0 m 1.00 s Definition Average distance. given mo = 10.5 m/s. Units: m Notes: Speer <- It is 2.3 F When eratic time, Like com (su car xc an If you use the same procedure to find the average velocities for intervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005- respectively. As At gets smaller and smaller, the average velocity closer and closer to 10.0 m/s. We conclude that the instanta velocity at time t = 1.0 s is 10.0 m/s. REFLECT As the time interval At approaches zero, the average ity in the interval is closer and closer to the limiting value 1- which we call the instantaneous velocity at time t = 1.00 s.E when we calculate an average velocity, we need to specify twa the beginning and end times of the interval-but for inst velocity at a particular time, we specify only that one time. Practice Problem: What is the cheetah's average speed c first second of the attack and (b) the first two seconds of Answers: (a) 5.00 m/s, (b) 10.0 m/s.
Transcribed Image Text:(a) The situation (b) Our sketch Vehicle ot Ax Uav,x= = At 2 We choose to place the origin at the vehicle. At time 12 = 2.00 s, the cheetah's position x2 is x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m. The displacement during this interval is SOLUTION SET UP Figure 2.10b shows the diagram we sketch. First we define a coordinate system, orienting it so that the cheetah runs in the +x direc- tion. We add the points we are interested in, the values we know, and the values we will need to find for parts (a) and (b). 40.0 m - 25.0 m 2.00 s-1.00 s We draw an axis. We point it in the direction the cheetah runs, so that our values will be positive. A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw. Cheetah Cheetah starts = 4-0 xo-20.0 m SOLVE Part (a): To find the displacement Ax, we first find the so that cheetah's positions (the values of x) at time t₁ = 1.00 s and at time 12=2.00 s by substituting the values of t into the given equation. At time = 1.00 s, the cheetah's position x₁ is x₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m. 0.02 15.0 m. → x₁ = ? 1₁-1.00 s We mark the initial positions of the cheetah and the antelope. (We won't use the antelope's position-but we don't know that yet.) Ax= x₂ = x₁ = 40.0 m - 25.0 m Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can now find the average velocity for that interval: 15.0 m = 15.0 m/s. 1.00 s Ax-? Vov.x => 807 X₂=? 1₂-2.00 s We're interested in the cheetah's motion between 1s and 2 s after it begins running. We place dots to represent those points. Antelope Uav,.x = Ax At Antelope 50.0 m x (m) x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m, 26.05 m 1.10 s 5 We add symbols for known and unknown quantities. We use subscripts 1 and 2 for the points at t=1s and f= 2 s. Part (c): The instantaneous velocity at 1.00 s is approximately (but exactly) equal to the average velocity in the interval from t₁ =- t₂ = 1.10 s (i.e., At = 0.10 s). At 1₂ = 1.10 s, 25.0 m 1.00 s Definition Average distance. given mo = 10.5 m/s. Units: m Notes: Speer <- It is 2.3 F When eratic time, Like com (su car xc an If you use the same procedure to find the average velocities for intervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005- respectively. As At gets smaller and smaller, the average velocity closer and closer to 10.0 m/s. We conclude that the instanta velocity at time t = 1.0 s is 10.0 m/s. REFLECT As the time interval At approaches zero, the average ity in the interval is closer and closer to the limiting value 1- which we call the instantaneous velocity at time t = 1.00 s.E when we calculate an average velocity, we need to specify twa the beginning and end times of the interval-but for inst velocity at a particular time, we specify only that one time. Practice Problem: What is the cheetah's average speed c first second of the attack and (b) the first two seconds of Answers: (a) 5.00 m/s, (b) 10.0 m/s.
Name: Page 35-Practice Problem 2.3: G# Page 34- Practice Problem 2.2: What is the cheetah's average speed over (a) the first second of the attack and (b) the first two seconds of the attack? Answers: (a) 5.00m/s, (b) 10.0m/s.
Transcribed Image Text:Name: Page 35-Practice Problem 2.3: G# Page 34- Practice Problem 2.2: What is the cheetah's average speed over (a) the first second of the attack and (b) the first two seconds of the attack? Answers: (a) 5.00m/s, (b) 10.0m/s.
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