4 1 Solve a b A race car slows from an initial speed of 100 mph to 50 mph in order to negotiate a tight turn After making the 90 turn the car accelerates back up to 100 mph in the same time it took to slow down 4 2 Solve a b A car drives up a hill over the top and down the other side at constant speed 4 3 Solve a b A ball rolls along a level table at 3 m s It rolls over the edge and falls 1 m to the floor How far away from the edge of the table does it land 4 4 Solve a The figure shows a motion diagram of a pendulum as it swings from one side to the other It s clear that the velocity at the lowest point is not zero The velocity vector at this point is tangent to the circle We can use the method of Tactics Box 1 3 to find the acceleration at the lowest point The acceleration is not zero Instead you can see that the acceleration vector points toward the center of the circle b The end of the arc is like the highest point of a ball tossed straight up The velocity is zero for an instant as the vector changes from pointing outward to pointing inward However the acceleration is not zero at this point cid 71 The velocity is changing at the end point and this requires an acceleration The motion diagram shows that v and thus is tangent to the circle at the end of the arc cid 71 a 4 5 Model The boat is treated as a particle whose motion is governed by constant acceleration kinematic equations in a plane Visualize Solve Resolving the acceleration into its x and y components we obtain i 0 80 m s sin 40 j 0 613 m s 2 2 i 0 514 m s 2 j cid 71 a 2 0 80 m s cos 40 cid 71 cid 71 a t v 0 1 cid 71 v 1 From the velocity equation cid 71 v 1 0 613 m s cid 71 The magnitude and direction of v 5 0 m s are i 2 t 0 i 0 514 m s 2 j 6 s 0 s 8 68 m s i 3 09 m s j v 8 68 m s 2 3 09 m s 9 21 m s 2 tan 1 tan 1 20 north of east v 1 v 1 y x 3 09 m s 8 68 m s Assess An increase of speed from 5 0 m s to 9 21 m s is reasonable 4 6 Solve cid 71 r j i 0 m 0 x and y a At t 0 s x 0 m and y 0 m or At t 4 s 0 mx and y 0 m or cid 71 r 0 i j 0 m In other words the particle is at the origin at both t 0 s and at t 4 s From the expressions for At t 0 s 2 m s j b At t 0 s is along cid 71 v cid 71 v 2 m s v At 4 s j or 90 south of t cid 71 v dx dt i dy dt 4 t i t 2 j m s v 8 3 m s j 2 m s 4 s t 23 t 2 j x At cid 71 v i 8 1 2 m s 8 m s tan x 14 north of 4 7 Visualize Refer to Figure EX4 7 Solve From the figure identify the following x 1 x 2 xv xv 2 1 0 m 2000 m 0 m s 200 m s 0 m y 1 y 2 1000 m yv 200 m s yv 2 1 100 m s The components of the acceleration can be found by applying 2 a s for the x and y directions Thus 2 v 2 2 v 1 a x 2 v 2 2 v x 1 x x 2 2 200 m s 2 2000 m 0 m s 0 m 2 10 00 m s 2 ya 2 100 m s 2 1000 m 0 m 200 m s 2 15 00 m s 2 Assess A time of 20 s is needed to change 0 m s to xv 2 200 m s at xa 2 10 m s This is the same time cid 71 a So i 10 00 2 j 15 00 m s v 1 y to v 2 y at a 2 15 m s y 1 xv needed to change 4 8 Model The puck is a particle and follows the constant acceleration kinematic equations of motion Visualize Please refer to Figure EX4 8 Solve 2 s cid 71 The angle made by the vector v the graphs give 30 cm s 16 cm s a At and t with the yv xv x axis can thus be found as tan 1 tan 62 above the x axis v y v x 1 30 cm s 16 cm s b After t 5 s the puck has traveled a distance given by x 1 x 0 5 s 0 v dt x 0 m area under v t x curve 1 2 40 cm s 5 s 100 cm y 1 y 0 v dt y 0 m area under v t y curve 30 cm s 5 s 150 cm 5 s 0 r 1 2 x 1 2 y 1 100 cm 2 150 cm 2 180 cm 4 9 Model Use the particle model for the puck Visualize Please refer to Figure EX4 9 Solve a Since the xv vs t and along the x and y axes The components of the puck s acceleration are yv vs t graphs are straight lines the puck is undergoing constant acceleration a x dv x dt v x t 10 m s 10 m s 10 s 0 s 2 0 m s 2 ya 10 m s 0 m s 10 s 0 s 1 0 m s 2 The magnitude of the acceleration is a a 2 x a 2 y 2 2 2 m s b The puck is undergoing constant acceleration in both the x and y directions Identify from the graphs xiv 10 m s yiv 0 m s Since the puck starts at the origin x i y i 0 m and set it 0 s Using kinematics The distance from the origin at time t is t 5 10 s 0 r 2 x y 2 The table below shows the values of x y and r at the times x 0 m 10 m s t 2 2 0 m s t 2 1 2 y 0 m 0 m s 2 1 0 m s t 2 1 2 0 s t 5 s 10 s x 0 m 25 m 0 m y 0 m 12 5 m 50 m …
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