HW1 - Solutions to Homework #1 Phys 125 Assignment Stage 0...

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Solutions to Homework #1 Phys 125 Assignment Stage 0 1.9. Solve: (a) Acceleration is found by the method of Tactics Box 1.3. Let 0 v r be the velocity vector between points 0 and 1 and be the velocity vector between points 1 and 2. 1 v r (b) Speed is greater than speed because more distance is covered in the same interval of time. 1 v 0 v 1.11. Solve: The acceleration vector at each location points directly toward the center of the Ferris wheel’s circular motion.
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1.43. Model: Represent Bruce and the puck as particles for the motion diagram. Visualize: 2.9. Solve: (a) The time for each of segment is and The average speed to the house is 1 50 mi/40 mph 5/4 hr t Δ= = 2 50 mi/60 mph 5/6hr. t = 100 mi 48 mph 5/6 hr 5/4 hr = + (b) Julie drives the distance Δ x 1 in time Δ t 1 at 40 mph. She then drives the distance Δ x 2 in time Δ t 2 at 60 mph. She spends the same amount of time at each speed, thus 12 1 2 1 /40 mph /60 mph (2/3) tt x x x x Δ=Δ⇒Δ ⇒Δ = Δ 2 But Δ x 1 + Δ x 2 = 100 miles, so (2/3) Δ x 2 + Δ x 2 = 100 miles. This means Δ x 2 = 60 miles and Δ x 1 = 40 miles. Thus, the times spent at each speed are Δ t 1 = 40 mi/40 mph = 1.00 hr and Δ t 2 = 60 mi/60 mph = 1.00 hr. The total time for her return trip is Δ t 1 + Δ t 2 = 2.00 hr. So, her average speed is 100 mi/2 hr = 50 mph. 2.11. Visualize: Please refer to Figure Ex2.11. The particle starts at x 0 = 10 m at t 0 = 0. Its velocity decreases as time increases, goes through zero at t = 3 s, and then reverses direction. Solve: (a) Using the equation x f = x 0 + area of the velocity graph between t i and t f , 2 s 3 s 4 s 3 s 10 m area of trapazoid between 0 s and 2 s 1 10 m (12 m/s 4 m/s)(2 s) 26 m 2 10 m area of triangle between 0 s and 3 s 1 10 m (12 m/s)(3 s) 28 m 2 area between 3 s and 4 s 1 28 m ( 4 m/s 2 x x xx =+ + = = )(1 s) 26 m = (b) The particle reverses direction at t = 3 s.
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2.23. Model: Represent the car as a particle. Visualize: Solve: Note that the problem “ends” at a turning point, where the car has an instantaneous speed of 0 m/s before rolling back down. The rolling back motion is not part of this problem. If we assume the car rolls without friction, then we have motion on a frictionless inclined plane with an accleration Constant acceleration kinematics gives 2 sin sin20 3.35 m/s . ag g θ =− °=− 2 2 22 2 0 10 0 1 1 2 (30 m/s) 2 ( ) 0 m /s 2 134 m ( 3 . 3 5 m / s ) v vv a xx v a x x a =+ − ⇒ =+ ⇒= − = = Notice how the two negatives canceled to give a positive value for x 1 . Assess: We must include the minus sign because the a r vector points down the slope, which is in the negative x -direction.
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HW1 - Solutions to Homework #1 Phys 125 Assignment Stage 0...

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