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- Lately, Jason has been taking drivers' ed so that he can get his license. However, he's also been taking physics, and so he wonders what his acceleration is, when he drives on a circular path in an empty parking lot near Peachtree Mall with diameter 78.0 m, if he is driving at a constant speed of 23.0 m/s. What is his acceleration? O 66.1 m/s² 13.6 m/s2 O 0.590 m/s² O 1.70 m/s² O 0 m/s²35. A person going for a walk follows the path shown in T Figure P3.35. The total trip consists of four straight-line paths. At the end of the walk, what is the person's resultant displacement measured from the starting point? Start 100 m 300 m End 200 m 30.0 150 m 60.0° Figure P3.35An object is fired vertically up from the edge of the top of a 80 m high building. It reaches the ground (at the bottom of the building) after 24 s. HowN Vong does it take the object to reach maximum height? Round the answer to three significant figures. 80 m
- Ans. s= 20 ft. 3. A particle travels along a straight line with a velocity v = (12 - 3r) m/s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t= 4 s, the displacement from 1 = 0 to t = 10 s, and the distance the particle travels during this time period. A, = -880 m ; Ans: a = -24 m/s? ; ST = 912 m 4. A particle travels along a straight line with a constant acceleration. When s = 4 ft, v = 3 ft/s and when s = 10 ft, v = 8 ft/s. Determine the velocity as a function of position.My question isn't how to solve the problem exactly. In fact, it's already been solved on this website. My question is about the acceleration. When I solve this problem myself, first I calculate the velocity by dividing 100m by 53s. I get 1.89m/s. Then I use that to find the acceleration using the equation vf = vi + at. That's 1.89/53 = 0.036m/s^2. That's not correct. The correct way to find the acceleration is to us the equation d = 1/2 at^2 and solve that way without taking the intermediate step of finding the velocity. Doing it that way, the acceleration is 0.0712m/s^2. My question is why you get a different result doing it the first way than you get doing it the second way.Report 1. From t = 0 to t = 3.44 min, a man stands still, and from t = 3.44 min to t = 6.88 min, he walks briskly in a straight line at a constant speed of 1.94 m/s. What is his average velocity vavg in the time interval 1.00 min to 4.44 min? O a. 1.940 m/s O b. 0.000 m/s O c. 0.970 m/s O d. 0.564 m/s
- 4. The acceleration of a particle is given by a = Avt where A = 2.0 m/s5/2. At t = 0, v = 7.5 m/s and x = 0. What is the displacement at t = 5.0 s? a =AVt %3D On nhiect has velocity Yo (t) = 31+1î 31+1i m/s at time t = 0.00 s. The acceleration of theStarting from the front door of a ranch house, you walk 60.0 mdue east to a windmill, turn around, and then slowly walk 40.0 m westto a bench, where you sit and watch the sunrise. It takes you 28.0 sto walk from the house to the windmill and then 36.0 s to walk fromthe windmill to the bench. For the entire trip from the front door to thebench, what are your average velocityA particle travels m a circular orbit of radius 10 m. Its speed is changing at a rate of 15.0m/s2 at an instant when its speed is 40.0 m/s. What is the magnitude of the acceleration of the particle?
- In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (69.0 m wide) and traveling 35.8 m/s off the takeoff ramp, he reached the other side. What was his launch angle?A football player runs from his own goal line to the opposing teams goal line, returning to the fifty-yard line, all in 18.0 s. Calculate (a) his average speed, and (b) the magnitude of his average velocity. (Sec Section 2.2)EXE A bus makes a trip according to the position-time graph shown in the drawing. What is the average velocity (magnitude and direction) of the bus during (a) segment A. (b) segment B. and (c) segment C? Express your answers in km/h. 50.0 (A) Number (by Number Number Pater m | +40.0 +30.0 +200 +100 Units Units 05 10 15/ 2.05 26 Time (