Ch 6 Physics for Scientists and Engineers

# Ch 6 Physics for Scientists and Engineers - 6.1 Model We...

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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: rr r r rrv tt a 10 0 10 1 2 2 =+ () +− ⇒+ =− ˆˆ 24 4 0 s3 s m 2 r r va i j From the kinematic equation for velocity: r vva 1 0 =+ − ⇒− 55 0 ij v a m/s 3 s 0 s r r ⇒= r r i j 0 3s m/s Substituting this form of r v 0 into the position equation, we find 2s ±m/s ±m 2 [] + aij a ij 44 ⇒= − r 22 m/s 2 Substituting r a back into either the velocity equation or the position equation gives r vi j 0 =−+ ( ) m/s. (b) Using the kinematic equation r r a 211 21 1 2 2 , 82 1 2 2 m m/s 5 s 3 s m/s 5 s 3 s 2 ( ) 22 16 m r 2 1 m/s m/s 5 s 3 s 2 ( ) 99 m/s Thus, the speed at t = 5 s is vv v xy 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 aij = () °+ ° 0.80 m/s cos40 0.80 m/s sin40 22 ˆˆ = + 0.613 m/s 0.514 m/s ij From the velocity equation rr r vva tt 10 1 0 =+ − , r vi i j 1 = + + [] 5.0 m/s 0.613 m/s 0.514 m/s 6 s 0 s ˆˆˆ = + 8.68 m/s 3.09 m/s The magnitude and direction of r v are v = + = 8.68 m/s 3.09 m/s 9.21 m/s 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.
6.3. Solve: (a) (b) At t = 0 s, x = 0 m and y = 0 m, or r ri j =+ () 00 m ˆˆ . At t = 4 s , x = 0 m and y = 0 m, or r j m . 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 =− +− 3 2 42 2 tt i tj m/s At t = 0 s, r vj 2 m / s ˆ , v = 2 m/s . At t = 4 s, r vij 82 m / s , v = 83 . 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 11 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: xx v d t x s 10 0 5 0 =+ = + m area under v x - t curve = () = 1 2 40 cm/s 5 s 100 cm yy v d t y s 0 5 0 m area under v y - t curve = = 30 cm/s 5 s 150 cm ⇒= + = + = rx y 2 1 2 100 cm 150 cm 180 cm 2 2
6.5. Model: The model rocket and the target will be treated as particles. Kinematic equations in two dimensions apply. Visualize: Solve: For the rocket, Newton’s second law along the y -direction is rr r r F F mg ma a m Fm g net R R RR 22 kg 15 N 0.8 kg 9.8 m/s m/s =− = ⇒= () =− [] = 11 08 895 .

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## This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

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Ch 6 Physics for Scientists and Engineers - 6.1 Model We...

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