Chapter06 - 6.1 Model We will assume motion under...

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6.1. Model: We will assume motion under constant-acceleration kinematics in a plane. Visualize: Instead of working with the components of position, velocity, and acceleration in the x and y directions, we will use the kinematic equations in vector form. Solve: (a) From the kinematic equation for position, after some manipulation: r r r r r r v t t a t t 1 0 0 1 0 1 2 1 0 2 = + ( ) + ( ) + ( ) = ( ) ( ) ˆ ˆ 2 4 4 0 s 3 s m 2 r r v a i j From the kinematic equation for velocity: r r r v v a t t 1 0 1 0 = + ( ) ( ) = + ( ) 5 5 0 ˆ ˆ i j v a m/s 3 s 0 s r r = − ( ) + ( ) r r v a i j 0 5 5 3 s m/s ˆ ˆ Substituting this form of r v 0 into the position equation, we find 2 s 3 s m/s 3 s m 2 ( ) ( ) + ( ) [ ] + ( ) = ( ) r r a i j a i j 5 5 4 4 ˆ ˆ ˆ ˆ = ( ) r a i j 2 2 ˆ ˆ m/s 2 Substituting r a back into either the velocity equation or the position equation gives r v i j 0 = − + ( ˆ ˆ ) m/s . (b) Using the kinematic equation r r r r r r v t t a t t 2 1 1 2 1 1 2 2 1 2 = + ( ) + ( ) , 8 2 5 5 2 2 1 2 2 ˆ ˆ ˆ ˆ ˆ ˆ i j i j i j ( ) + ( ) [ ] ( ) + ( ) [ ] ( ) m m/s 5 s 3 s m/s 5 s 3 s 2 = ( ˆ ˆ ) 22 16 i j m r r r v v a t t 2 1 2 1 = + ( ) = ( ) + ( ) [ ] ( ) 5 5 2 2 ˆ ˆ ˆ ˆ i j i j m/s m/s 5 s 3 s 2 = ( ˆ ˆ ) 9 9 i j m/s Thus, the speed at t = 5 s is v v v x y 2 2 2 2 2 2 2 9 9 12 7 = ( ) + ( ) = ( ) + − ( ) = m/s m/s . .
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6.2. 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 r a i j = ( ) ° + ( ) ° 0.80 m/s cos40 0.80 m/s sin40 2 2 ˆ ˆ = ( ) + ( ) 0.613 m/s 0.514 m/s 2 2 ˆ ˆ i j From the velocity equation r r r v v a t t 1 0 1 0 = + ( ) , r v i i j 1 = ( ) + ( ) + ( ) [ ] ( ) 5.0 m/s 0.613 m/s 0.514 m/s 6 s 0 s 2 2 ˆ ˆ ˆ = ( ) + ( ) 8.68 m/s 3.09 m/s ˆ ˆ i j The magnitude and direction of r v are v = ( ) + ( ) = 8.68 m/s 3.09 m/s 9.21 m/s 2 2 θ = = = ° tan tan 1 1 1 1 20 v v y x 3.09 m/s 8.68 m/s north of east Assess: An increase of speed from 5.0 m/s to 9.21 m/s is reasonable.
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6.3. Solve: (a) (b) At t = 0 s, x = 0 m and y = 0 m, or r r i j = + ( ) 0 0 m ˆ ˆ . At t = 4 s , x = 0 m and y = 0 m, or r r i j = + ( ) 0 0 m ˆ ˆ . In other words, the particle is at the origin at both t = 0 s and at t = 4 s . From the expressions for x and y , r v dx dt i dy dt j = + ˆ ˆ = + ( ) 3 2 4 2 2 t t i t j ˆ ˆ m/s At t = 0 s, r v j = − 2 m/s ˆ , v = 2 m/s . At t = 4 s, r v i j = + ( ) 8 2 m/s ˆ ˆ , v = 8 3 . m/s . (c) At t = 0 s, r v is along ˆ j , or 90 ° south of + x . At t = 4 s, θ = = ° tan 1 14 2 m/s 8 m/s north of + x
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6.4. Model: The puck is a particle and follows the constant-acceleration kinematic equations of motion. Visualize: Please refer to Figure Ex6.4. Solve: (a) At t = 2 s, the graphs give v x = 16 cm/s and v y = 30 cm/s. The angle made by the vector r v with the x -axis can thus be found as θ = = = ° tan tan 1 1 62 v v y x 30 cm/s 16 cm/s above the x -axis (b) After t = 5 s, the puck has traveled a distance given by: x x v dt x s 1 0 0 5 0 = + = + m area under v x - t curve = ( ) ( ) = 1 2 40 cm/s 5 s 100 cm y y v dt y s 1 0 0 5 0 = + = + m area under v y - t curve = ( ) ( ) = 30 cm/s 5 s 150 cm = + = ( ) + ( ) = r x y 1 1 2 1 2 100 cm 150 cm 180 cm 2 2
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6.5.
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