Chapter 2

Chapter 2 - 2.1. Model: We will consider the car to be a...

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2.1. Model: We will consider the car to be a particle that occupies a single point in space. Visualize: Solve: Since the velocity is constant, we have fi . x x xvt = +Δ Using the above values, we get 1 0m (10 m/s)(45s) 450 m x =+ = Assess: 10 m/s 22 mph and implies a speed of 0.4 miles per minute. A displacement of 450 m in 45 s is reasonable and expected.
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2.2. Model: We will consider Larry to be a particle. Visualize: Solve: Since Larry’s speed is constant, we can use the following equation to calculate the velocities: fi s s s v tt = (a) For the interval from the house to the lamppost: 1 200 yd 600 yd 200 yd/min 9:07 9:05 v == For the interval from the lamppost to the tree: 2 1200 yd 200 yd 333 yd/min 9:10 9:07 v + (b) For the average velocity for the entire run: avg 1200 yd 600 yd 120 yd/min 9:10 9:05 v +
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2.3. Model: Cars will be treated by the particle model. Visualize: Solve: Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows: 10 x xx vt t ttt v Δ −− == = + Δ− Using the known values identified in the pictorial representation, we find: Alan 1 Alan 0 Alan 1 Alan 0 Beth 1 Beth 0 Beth 1 Beth 0 400 mile 8:00 AM 8:00 AM 8 hr 4:00 PM 50 miles/hour 400 mile 9:00 AM 9:00 AM 6.67 hr 3:40 PM 60 miles/hour tt v v =+ = + = + = = + = + = (a) Beth arrives first. (b) Beth has to wait Alan 1 Beth 1 20 minutes −= for Alan. Assess: Times of the order of 7 or 8 hours are reasonable in the present problem.
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2.4. Solve: (a) The time for each segment is 1 50 mi/40 mph 5/4 hr t Δ= = and 2 50 mi/60 mph 5/6hr. t = The average speed to the house is 100 mi 48 mph 5/6 h 5/4 h = + (b) Julie drives the distance 1 x Δ in time 1 t Δ at 40 mph. She then drives the distance 2 x Δ in time 2 t Δ at 60 mph. She spends the same amount of time at each speed, thus 12 1 2 1 2 /40 mph /60 mph (2/3) tt x x x x Δ=Δ⇒Δ ⇒Δ= Δ But 100 miles, xx Δ+ so 22 (2/3) 100 miles. This means 2 60 miles x and 1 40 miles. x Thus, the times spent at each speed are 1 40 mi/40 mph 1.00 h t = and 2 60 mi/60 mph 1.00 h. t = The total time for her return trip is 2.00 h. Δ+Δ= So, her average speed is 100 mi/2 h 50 mph. =
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2.5. Model: The bicyclist is a particle. Visualize: Please refer to Figure EX2.5. Solve: The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at t = 10 s is 100 m 50 m 2.5 m/s 20 s s v t Δ− == = Δ The slope at 25 s t = is 100 m 100 m 0 m/s 10 s v The slope at 35 s t = is 0 m 100 m 10 m/s 10 s v
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2.6. Visualize: Please refer to Figure EX2.6. Solve: (a) We can obtain the values for the velocity-versus-time graph from the equation /. vs t =Δ Δ (b) There is only one turning point. At 3 s t = the velocity changes from 5 m/s + to 20 m/s, thus reversing the direction of motion. At 1 s, t = there is an abrupt change in motion from rest to 5 m/s, + but there is no reversal in motion.
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2.7. Visualize: Please refer to Figure EX2.7. The particle starts at 0 10 m x = at 0 0. t = Its velocity is initially in the x direction. The speed decreases as time increases during the first second, is zero at 1 s, t = and then increases after the particle reverses direction.
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Chapter 2 - 2.1. Model: We will consider the car to be a...

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