Practice problem 2.2 What is the cheetahs 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) 10.0 m/sBoth photos have background info to solve the practice problem. 34 CHAPTER 2 Motion Along a Straight Line(a) The situation(b) Our sketchUav,x=AxAtVehicle=VehicleofAt time = 2.00 s, the cheetah's position x₂ is1₂x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m.The displacement during this interval isSOLUTIONSET UP Figure 2.10b shows the diagram we sketch. First we define acoordinate 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 thevalues we will need to find for parts (a) and (b).40.0 m - 25.0 m2.00 s-1.00 sCheetahWe draw an axis.We point it in thedirection thecheetah runs, sothat our valueswill be positive.A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw.SOLVE Part (a): To find the displacement Ax, we first find thecheetah's positions (the values of x) at time t₁ = 1.00 s and at time12= 2.00 s by substituting the values of t into the given equation. Attime = 1.00 s, the cheetah's position x₁ isx₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m.CheetahstartsXo-20.0m+-02 We choose 3 We mark the initialto place theorigin at thevehicle.A.x = x₂x₁ = 40.0 m 25.0 m = 15.0 m.Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can nowfind the average velocity for that interval:=V₁x = ?X₁₂-?1₁-1.00 s15.0 m= 15.0 m/s.1.00 spositions of the cheetahand the antelope. (Wewon't use the antelope'sposition-but we don'tknow that yet.)Ax=?Vav,x=?so thatX₂=?t₂ = 2.00 sWe're interested in thecheetah's motionbetween 1s and 2 s afterit begins running. Weplace dots to representthose points.Uav,x=AntelopeAxAtAntelopePart (c): The instantaneous velocity at 1.00 s is approximately (but notexactly) equal to the average velocity in the interval from ₁= 1.00 s to12 1.10 s (i.e., Ar= 0.10 s). At t₂ = 1.10 s,x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m,26.05 m1.10 s50.0 m→x (m)65 We add symbols forknown and unknownquantities. We usesubscripts 1 and 2 forthe points at f= 1 sand r = 2 s.25.0 m1.00 s= 10.5 m/s.If you use the same procedure to find the average velocities for timeintervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005 m/s,respectively. As At gets smaller and smaller, the average velocity getscloser and closer to 10.0 m/s. We conclude that the instantaneousvelocity 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 thatwhen we calculate an average velocity, we need to specify two times-the beginning and end times of the interval-but for instantaneousvelocity at a particular time, we specify only that one time.Practice Problem: What is the cheetah's average speed over (a) thfirst second of the attack and (b) the first two seconds of the attackAnswers: (a) 5.00 m/s, (b) 10.0 m/s. avelocity hom graphs UI position versus timeFigure 2.9 shows graphs of an object's position along the x axis versustime for four separate trials. Which of these statements is or are correct?A FIGURE 2.9XXA. The velocity is greater in trial 2 than in trial 3.B. The velocity is not constant during trial 1.C. During trial 4, the object changes direction as it passes through thepoint x = 0.D. The velocity is constant during trials 2, 3, and 4.(Remember that velocity means "x component of velocity.")SOLUTION The graph of the motion during trial I has a changingslope and therefore v, isn't constant. Trials 2, 3, and 4 all have graphswith constant slope and thus correspond to motion with constant veloc-ity. (The velocity is zero during trial 2, as indicated by the slope of zero,but this value is still constant.) Trial 3 has a positive velocity, while thevelocity in trial 4 is negative. Thus, answers B and D are correct.EXAMPLE 2.2 Average and instantaneous velocitiesHere we will look at the difference between average and instantaneous velocities via a specific example ofa cheetah chasing its prey. The cheetah is crouched in ambush 20.0 m to the east of an observer's vehicle,as shown in Figure 2.10a. At time t = 0, the cheetah charges an antelope in a clearing 50.0 m east of theobserver. The cheetah runs along a straight line; the observer estimates that, during the first 2.00 s of theattack, the cheetah's coordinate x varies with time t according to the equationx = 20.0 m+ (5.00 m/s²)t².(a) Find the displacement of the cheetah during the interval between t₁ = 1.00 s and t₂ = 2.00 s. (b) Findthe average velocity during this time interval. (c) Estimate the instantaneous velocity at time t₁ = 1.00 s bytaking At = 0.10 s.Video Tutor SYOLC

Question

Practice problem 2.2 What is the cheetahs 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) 10.0 m/sBoth photos have background info to solve the practice problem. 34 CHAPTER 2 Motion Along a Straight Line(a) The situation(b) Our sketchUav,x=AxAtVehicle=VehicleofAt time = 2.00 s, the cheetah's position x₂ is1₂x₂ = 20.0 m+ (5.00 m/s²) (2.00 s)² = 40.0 m.The displacement during this interval isSOLUTIONSET UP Figure 2.10b shows the diagram we sketch. First we define acoordinate 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 thevalues we will need to find for parts (a) and (b).40.0 m - 25.0 m2.00 s-1.00 sCheetahWe draw an axis.We point it in thedirection thecheetah runs, sothat our valueswill be positive.A FIGURE 2.10 (a) The situation for this problem; (b) the sketch we draw.SOLVE Part (a): To find the displacement Ax, we first find thecheetah's positions (the values of x) at time t₁ = 1.00 s and at time12= 2.00 s by substituting the values of t into the given equation. Attime = 1.00 s, the cheetah's position x₁ isx₁ = 20.0 m+ (5.00 m/s²) (1.00 s)² = 25.0 m.CheetahstartsXo-20.0m+-02 We choose 3 We mark the initialto place theorigin at thevehicle.A.x = x₂x₁ = 40.0 m 25.0 m = 15.0 m.Part (b): Knowing the displacement from 1.00 s to 2.00 s, we can nowfind the average velocity for that interval:=V₁x = ?X₁₂-?1₁-1.00 s15.0 m= 15.0 m/s.1.00 spositions of the cheetahand the antelope. (Wewon't use the antelope'sposition-but we don'tknow that yet.)Ax=?Vav,x=?so thatX₂=?t₂ = 2.00 sWe're interested in thecheetah's motionbetween 1s and 2 s afterit begins running. Weplace dots to representthose points.Uav,x=AntelopeAxAtAntelopePart (c): The instantaneous velocity at 1.00 s is approximately (but notexactly) equal to the average velocity in the interval from ₁= 1.00 s to12 1.10 s (i.e., Ar= 0.10 s). At t₂ = 1.10 s,x₂ = 20.0 m+ (5.00 m/s) (1.10 s)² = 26.05 m,26.05 m1.10 s50.0 m→x (m)65 We add symbols forknown and unknownquantities. We usesubscripts 1 and 2 forthe points at f= 1 sand r = 2 s.25.0 m1.00 s= 10.5 m/s.If you use the same procedure to find the average velocities for timeintervals of 0.01 s and 0.001 s, you get 10.05 m/s and 10.005 m/s,respectively. As At gets smaller and smaller, the average velocity getscloser and closer to 10.0 m/s. We conclude that the instantaneousvelocity 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 thatwhen we calculate an average velocity, we need to specify two times-the beginning and end times of the interval-but for instantaneousvelocity at a particular time, we specify only that one time.Practice Problem: What is the cheetah's average speed over (a) thfirst second of the attack and (b) the first two seconds of the attackAnswers: (a) 5.00 m/s, (b) 10.0 m/s. avelocity hom graphs UI position versus timeFigure 2.9 shows graphs of an object's position along the x axis versustime for four separate trials. Which of these statements is or are correct?A FIGURE 2.9XXA. The velocity is greater in trial 2 than in trial 3.B. The velocity is not constant during trial 1.C. During trial 4, the object changes direction as it passes through thepoint x = 0.D. The velocity is constant during trials 2, 3, and 4.(Remember that velocity means &#34;x component of velocity.&#34;)SOLUTION The graph of the motion during trial I has a changingslope and therefore v, isn't constant. Trials 2, 3, and 4 all have graphswith constant slope and thus correspond to motion with constant veloc-ity. (The velocity is zero during trial 2, as indicated by the slope of zero,but this value is still constant.) Trial 3 has a positive velocity, while thevelocity in trial 4 is negative. Thus, answers B and D are correct.EXAMPLE 2.2 Average and instantaneous velocitiesHere we will look at the difference between average and instantaneous velocities via a specific example ofa cheetah chasing its prey. The cheetah is crouched in ambush 20.0 m to the east of an observer's vehicle,as shown in Figure 2.10a. At time t = 0, the cheetah charges an antelope in a clearing 50.0 m east of theobserver. The cheetah runs along a straight line; the observer estimates that, during the first 2.00 s of theattack, the cheetah's coordinate x varies with time t according to the equationx = 20.0 m+ (5.00 m/s²)t².(a) Find the displacement of the cheetah during the interval between t₁ = 1.00 s and t₂ = 2.00 s. (b) Findthe average velocity during this time interval. (c) Estimate the instantaneous velocity at time t₁ = 1.00 s bytaking At = 0.10 s.Video Tutor SYOLC

Accepted Answer

The cheetah travels from one position to the other in a given time period. It is stated that the…