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 x f x i x Using the above values we get v t x 1 0 m 10 m s 45 s 450 m 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 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 v s s f t f s i t i v 1 200 yd 600 yd 9 07 9 05 200 yd min v 2 1200 yd 200 yd 9 10 9 07 333 yd min a For the interval from the house to the lamppost For the interval from the lamppost to the tree b For the average velocity for the entire run v avg 1200 yd 600 yd 9 10 9 05 120 yd min 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 x 0 t t 1 x t v x 0 x 1 v t 0 x 1 t 1 0 Using the known values identified in the pictorial representation we find t t t Alan 1 Alan 0 x Alan 1 x Alan 0 8 00 AM t Beth 1 Beth 0 x Beth 1 x Beth 0 9 00 AM v v 400 mile 50 miles hour 400 mile 60 miles hour 8 00 AM 8 hr 4 00 PM 9 00 AM 6 67 hr 3 40 PM a Beth arrives first t 20 minutes b Beth has to wait Alan 1 Assess Times of the order of 7 or 8 hours are reasonable in the present problem for Alan t Beth 1 2 4 Solve The average speed to the house is a The time for each segment is t 1 50 mi 40 mph 5 4 hr and t 2 50 mi 60 mph 5 6hr 100 mi 5 6 h 5 4 h 48 mph b Julie drives the distance She spends the same amount of time at each speed thus x 2 t 2 1x in time 40 mph x 1 t 1 60 mph x 1 2 3 x 2 1t at 40 mph She then drives the distance 2x in time 2t at 60 mph But 2 100 miles x x 1 so x 2 100 miles This means the times spent at each speed are 40 mi 40 mph 1 00 h and t 2 60 miles x 2 60 mi 60 mph 1 00 h x 1 and 40 miles Thus The total time t t 1 2 So her average speed is 100 mi 2 h 50 mph 2 3 x 2 t 1 2 00 h for her return trip is 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 The slope at t 25 s is The slope at t 35 s is v s t 100 m 50 m 20 s 2 5 m s v 100 m 100 m 10 s 0 m s v 0 m 100 m 10 s 10 m s 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 v s t t there is an abrupt change in motion from rest to 5 m s to 20 m s the velocity changes from 5 m s thus reversing the but there is no reversal 3 s b There is only one turning point At direction of motion At in motion 1 s t 2 7 Visualize Please refer to Figure EX2 7 The particle starts at initially x direction The speed decreases as time increases during the first second is zero at after the particle reverses direction Solve b Using the equation when vx changes sign area of the velocity graph between 1t and f t a The particle reverses direction at x 0 x 0 10 m 1 s t in x f x 2 s 10 m area of triangle between 0 s and 1 s area of triangle between 1 s and 2 s at 0 t t 1 s 0 Its velocity is the and then increases 10 m 4 m s 1 s 4 m s 1 s 10 m 1 2 x 3 s 10 m area of trapazoid between 2 s and 3 s 10 m 4 m s 8 m s 3 s 2 s 16 m x 4 s x 3 s area between 3 s and 4 s 16 m 8 m s 12 m s 1 s 26 m 1 2 1 2 1 2 2 8 Visualize Please refer to Figure EX2 8 2 s Solve A constant velocity from t in velocity between t 4 s and to 2 s 0 s t t means zero acceleration On the other hand a linear increase implies a constant positive acceleration 2 9 Visualize Please refer to Figure EX2 9 Solve a The acceleration of the train at a 8 s 2 m s 0 5 m s 2 m s t 2 b 3 0 s is the slope of the v vs t graph at t 3 s Thus 2 10 Visualize Please refer to Figure EX2 10 Solve the position of the particle is a At 2 0 s t x 2 s t 2 0 m area under velocity graph from t 0 s to 2 0 s 2 0 m 4 m s 2 0 s 6 m 1 2 b From the graph itself at c The acceleration is t 2 0 s v 4 m s a x v x t v x f v i x t 6 m s 0 m s 3 s 2 m s 2 2 11 Visualize Please refer to Figure EX2 11 Solve a Using the equation x f x i area under the velocity versus time graph between it and ft x t at 1 s x at t 0 s area between t 0 s and t 1 s 2 0 m 4 m s 1 s 6 m Reading from the velocity versus time graph xv at area between at b t x 2 0 m 4 m s 2 s 2 m s 1 s 3 0 s at t 0 s x t 1 s xa Also 1 2 2 m s 1 s 13 0 m 4 m s and 3 s t 0 s t slope v t 2 0 m s Reading from the graph xv t 3 s 2 m s …
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