ssm_ch04 - Chapter 4 Student Solutions Manual 11. We apply...

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Chapter 4 – Student Solutions Manual 11. We apply Eq. 4-10 and Eq. 4-16. (a) Taking the derivative of the position vector with respect to time, we have, in SI units (m/s), 2 ˆˆ = (i + 4 j + k) = 8 j + k . d vt t t dt G ˆ (b) Taking another derivative with respect to time leads to, in SI units (m/s 2 ), ˆˆ ˆ = (8 j + k) = 8 j . d at dt G 17. Constant acceleration in both directions ( x and y ) allows us to use Table 2-1 for the motion along each direction. This can be handled individually (for Δ x and Δ y ) or together with the unit-vector notation (for Δ r ). Where units are not shown, SI units are to be understood. (a) The velocity of the particle at any time t is given by G G G vv a t = + 0 , where is the initial velocity and is the (constant) acceleration. The x component is v G v 0 G a x = v 0 x + a x t = 3.00 – 1.00 t , and the y component is v y = v 0y + a y t = –0.500 t since v 0y = 0. When the particle reaches its maximum x coordinate at t = t m , we must have v x = 0. Therefore, 3.00 – 1.00 t m = 0 or t m = 3.00 s. The y component of the velocity at this time is v y = 0 – 0.500(3.00) = –1.50 m/s; this is the only nonzero component of G v at t m . (b) Since it started at the origin, the coordinates of the particle at any time t are given by GG G rv t a t =+ 0 1 2 2 . At t = t m this becomes () ( ) 2 1 ˆ ˆ 3.00i 3.00 1.00i 0.50 j 3.00 (4.50i 2.25 j) m. 2 r = G ˆ 29. The initial velocity has no vertical component — only an x component equal to +2.00 m/s. Also, y 0 = +10.0 m if the water surface is established as y = 0. (a) x x 0 = v x t readily yields x x 0 = 1.60 m. (b) Using yy v t g t y −= 00 1 2 2 , we obtain y = 6.86 m when t = 0.800 s and v 0 y =0.
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(c) Using the fact that y = 0 and y 0 = 10.0, the equation yy v t g t y −= 00 1 2 2 leads to 2 2(10.0 m)/9.80 m/s 1.43 s t == . During this time, the x -displacement of the diver is x x 0 = (2.00 m/s)(1.43 s) = 2.86 m. 31. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at ground level directly below the release point. We write θ 0 = –37.0° for the angle measured from + x , since the angle given in the problem is measured from the – y direction. We note that the initial speed of the projectile is the plane’s speed at the moment of release. (a) We use Eq. 4-22 to find v 0 : 2 22 0 0 11 ( sin ) 0 730 m sin( 37.0 )(5.00 s) (9.80 m/s )(5.00 s) t v = ° which yields v 0 = 202 m/s. (b) The horizontal distance traveled is x = v 0 t cos 0 = (202 m/s)(5.00 s)cos(–37.0°) = 806 m. (c) The x component of the velocity (just before impact) is v x = v 0 cos 0 = (202 m/s)cos(–37.0°) = 161 m/s. (d) The y component of the velocity (just before impact) is v y = v 0 sin 0 gt = (202 m/s) sin (–37.0°) – (9.80 m/s 2 )(5.00 s) = –171 m/s. 39. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet as it begins projectile motion in the sense of § 4-5), and we let 0 be the firing angle. If the target is a distance d away, then its coordinates are x = d, y = 0.
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This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.

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ssm_ch04 - Chapter 4 Student Solutions Manual 11. We apply...

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