Chapter 4

Chapter 4 - 4.1. Solve: (a) (b) A race car slows from an...

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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.
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4.2. Solve: (a) (b) A car drives up a hill, over the top, and down the other side at constant speed.
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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?
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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. The velocity is changing at the end point, and this requires an acceleration. The motion diagram shows that , v Δ G and thus , a G is tangent to the circle at the end of the arc.
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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 () ( ) ( ) ( ) 22 2 2 ˆˆ ˆ ˆ 0.80 m/s cos40 0.80 m/s sin 40 0.613 m/s 0.514 m/s aij i j = + G From the velocity equation ( ) 10 vva tt =+ − GGG , ( ) ( ) ( ) ( ) ( ) 1 ˆ ˆ ˆ 5.0 m/s 0.613 m/s 0.514 m/s 6 s 0 s 8.68 m/s 3.09 m/s vi i j i j ⎡⎤ =+ + = + ⎣⎦ G The magnitude and direction of v G are 8.68 m/s 3.09 m/s 9.21 m/s v = 1 11 1 3.09 m/s tan tan 20 north of east 8.68 m/s y x v v θ −− ⎛⎞ == = ° ⎜⎟ ⎝⎠ Assess: An increase of speed from 5.0 m/s to 9.21 m/s is reasonable.
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4.6. Solve: (a) At 0 s, t = 0 m x = and 0 m, y = or ˆˆ (0 0 ) m. ri j =+ G At 4 s, t = x = and 0 m, y = or (0 0 ) m. j G In other words, the particle is at the origin at both 0 s t = and at 4 s. t = From the expressions for x and y , () 2 3 ˆ ˆ 42 m / s 2 dx dy vi j t t i tj dt dt ⎛⎞ =+ = −+ ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ G At 0 s, t = ˆ 2 m /s , vj =− G 2 m/s. v = At t = ( ) 82 m / s , j G 8.3 m/s. v = (b) At 0 s, t = v G is along ˆ j , or 90 ° south of . x + At t = 1 2 m/s tan 14 north of + 8 m/s x θ == °
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4.7. Visualize: Refer to Figure EX4.7. Solve: From the figure, identify the following: 1 0 m x = 1 0 m y = 2 2000 m x = 2 1000 m y = 1 0 m/s x v = 1 200 m/s y v = 2 200 m/s x v = 2 100 m/s y v =− The components of the acceleration can be found by applying 22 21 2 vv a s = for the x and y directions. Thus () ( ) 2 200 m/s 0 m/s 10.00 m/s 2 2 2000 m 0 m xx x vv a x == = Δ− ( ) 2 100 m/s 200 m/s 15.00 m/s 2 1000 m 0 m y a −− So 2 ˆˆ (10.00 15.00 ) m/s .
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This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 4 - 4.1. Solve: (a) (b) A race car slows from an...

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