(a) The situation (b) Our sketch Vehicle Uav, x Vehicle At time = 2.00 s, the cheetah's position x₂ is The displacement during this interval is x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m. We choose to place the origin at the vehicle. We draw an axis. We point it in the direction the cheetah runs, so that our values will be positive. (3) 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.) A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw. 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). Ax=x₂x₁ = 40.0 m 25.0 m = 15.0 m. - Δ.r 40.0 m 25.0 m Ar 2.00 s-1.00 s Cheetah 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 f 1.00 s, the cheetah's position x, is x₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m. Cheetah starts Xo-20.0m +-0 15.0 m 1.00 s Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can now find the average velocity for that interval: VIX - ? x₁-? 15.0 m/s. ₁-1.00 s Ax-? Vav,x=? X2=? 1₂ -2.00 s Antelope We're interested in the cheetah's motion between 1s and 2 s after it begins running. We place dots to represent those points. Ax At Antelope 26.05 m 1.10 s 50.0 m Part (c): The instantaneous velocity at 1.00 s is approximately exactly) equal to the average velocity in the interval from ₁ t₂ = 1.10 s (i.e., At = 0.10 s). At t = 1.10 s, x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m, →x (m) 6 We add symbols for known and unknown quantities. We use subscripts 1 and 2 for the points at ris and r = 2 s. 25.0 m 1.00 s (but not 1.00 s to 10.5 m/s. Uav,x= 8.53 If you use the same procedure to find the average velocities for time intervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005 m/s, respectively. As Ar gets smaller and smaller, the average velocity gets closer and closer to 10.0 m/s. We conclude that the instantaneous velocity at time t 1.0 s is 10.0 m/s. REFLECT As the time interval Ar approaches zero, the average veloc ity in the interval is closer and closer to the limiting value 10.0 m/s which we call the instantaneous velocity at time t = 1.00 s. Note tha when we calculate an average velocity, we need to specify two times- the beginning and end times of the interval-but for instantaneo velocity at a particular time, we specify only that one time. Practice Problem: What the cheetah's average speed over (a) first second of the attack and (b) the first two seconds of the attac Answers: (a) 5.00 m/s, (b) 10.0 m/s.