ch3 - 3.3.(a) IDENTIFY and SET UP: From r we can calculate...

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3.3.(a) IDENTIFY and SET UP: From we can calculate x and y for any t . r G Then use Eq.(3.2), in component form. EXECUTE: ( ) () 22 ˆˆ 4.0 cm 2.5 cm/s 5.0 cm/s tt ⎡⎤ =+ + ⎣⎦ ri j G At 0, t = ˆ 4.0 cm . = G At 2.0 s, t = ( 14.0 cm 10.0 cm . ) j G av 10.0 cm 5.0 cm/s. 2.0 s x x v t Δ == = Δ av 10.0 cm 5.0 cm/s. 2.0 s y y v t Δ = Δ ()() av av av 7.1 cm/s xy vvv = ( ) av av tan 1.00 y x v v α 45 . θ = ° Figure 3.3a EVALUATE: Both x and y increase, so av v G is in the 1st quadrant. (b) IDENTIFY and SET UP: Calculate r G by taking the time derivative of () . t r G EXECUTE: 2 5.0 cm/s 5.0 cm/s d t dt + r vi j G G 0: t = 0, x v = 5.0 cm/s; y v = 5.0 cm/s v = and 90 = ° 1.0 s: t = 5.0 cm/s, x v = 5.0 cm/s; y v = 7.1 cm/s v = and 45 = ° 2.0 s: t = 10.0 cm/s, x v = 5.0 cm/s; y v = 11 cm/s v = and 27 = ° (c) The trajectory is a graph of y versus x . 4.0 cm (2.5 cm/s ) , x t (5.0 cm/s) yt = For values of t between 0 and 2.0 s, calculate x and y and plot y versus x . Figure 3.3b E VALUATE : The sketch shows that the instantaneous velocity at any t is tangent to the trajectory.
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3.7.IDENTIFY and SET UP: Use Eqs.(3.4) and (3.12) to find , x v , y v , x a and y a as functions of time. The magnitude and direction of r and G a G can be found once we know their components. EXECUTE: (a) Calculate x and y for t values in the range 0 to 2.0 s and plot y versus x . The results are given in Figure 3.7a. Figure 3.7a (b) x dx v dt α == 2 y dy vt dt β 0 x y dv a dt 2 y y dv a dt Thus ˆˆ 2 at =− vi j G ˆ 2 a j G (c) velocity : At 2.0 s, t = 2.4 m/s, x v = 2 2(1.2 m/s )(2.0 s) 4.8 m/s y v 22 5.4 m/s xy vv v =+ = 4.8 m/s tan 2.00 2.4 m/s y x v v = 63.4 360 297 = −° Figure 3.7b acceleration : At 2.0 s, t = 0, x a = 2(1.2 m/s ) 2.4 m/s y a 2.4 m/s aa a = 2 2 2.4 m/s tan 0 y x a a = 270 Figure 3.7c
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EVALUATE: (d) a G has a component in the same direction as a & , v G so we know that v is increasing (the bird is speeding up.) a G also has a component a perpendicular to so that the direction of v , v G G is changing; the bird is turning toward the -direction y (toward the right) Figure 3.7d v G is always tangent to the path; v G at 2.0 s t = shown in part (c) is tangent to the path at this t , conforming to this general rule. a G is constant and in the -direction; y the direction of is turning toward the v G -direction. y Figure 3.8a-c 3.9. IDENTIFY: The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal. SET UP: Take the positive y -direction to be upward. Take the origin of coordinates at the initial position of the book, at the point where it leaves the table top. x -component : 0, x a = 0 1.10 m/s, x v = 0.350 s t = y -component : 2 9.80 m/s , y a =− 0 0, y v = 0.350 s t = Figure 3.9a Use constant acceleration equations for the x and y components of the motion, with and . 0 x a = y ag EXECUTE: (a) 0 ? yy −= 22 2 11 00 0 ( 9.80 m/s )(0.350 s) 0.600 m. yy v t a t −= + =+− = The table top is 0.600 m above the floor. (b) 0 ? xx 2 1 2 (1.10 m/s)(0.350 s) 0 0.358 m. xx v t = += (c) (The x -component of the velocity is constant, since 0 1.10 m/s x vv a t =+= 0.) x a = 2 0 0 ( 9.80 m/s )(0.350 s) 3.43 m/s y t + =
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22 3.60 m/s xy vv v =+ = 3.43 m/s tan 3.118 1.10 m/s y x v v α == = 72.2 = −° Direction of v G is 72.2 below the horizontal ° Figure 3.9b (d) The graphs are given in Figure 3.9c Figure 3.9c EVALUATE: In the x -direction, 0 x a = and x v is constant. In the y -direction, and 2 9.80 m/s y a =− y v
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This note was uploaded on 04/29/2008 for the course PH 1110 taught by Professor Kiel during the Fall '08 term at WPI.

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ch3 - 3.3.(a) IDENTIFY and SET UP: From r we can calculate...

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