Sol-Ch04 - Exercises 1. (a) v = (6t – 2) i – 3t 2 j =...

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Unformatted text preview: Exercises 1. (a) v = (6t – 2) i – 3t 2 j = 10i – 12 j m / s; (b) a = 6i – 6tj = 6i – 24 j m / s 2 , (c) a a v = 6i –12 j m / s 2 . 2. (a) 0.5i – 1.5 j + 2k m / s; (b) v 1 = v 2 – a Δ t = 40i – 10 j – 2k m / s 3. (a) v a v = 4i + 7 j m / s ; (b) v i = 12i + 9 j and v f = 18i + 24 j ; so a a v = 1.2i + 3 j m / s 2 . 4. (a) 350 m; (b) 250 m, 37° N of E; (c) 10 m / s, 37° N of E; (d) 14 m / s ; (e) a av = (10 j – 20i ) / 25 = –0.8i + 0.4 j m / s 2 5. (a) Δ r = (0.707Ri + 0.707R j ) – R j = –09i – 0.37 j m; (b) Δ r = (0.707Ri – 0.707R j ) – R j = 0.3i – 0.71 j m / s; (c) Δ v = –(2) 1/2 i , a a v = Δ v / Δ t = –0.71i m / s 2 6. The 5810 km distance is along the curved surface of the earth. The angular displacement is Δ θ = 5810 km / 6370 km = 0.912 rad. The magnitude of the displacement is the length of the straight line joining the initial and final points :2 tan(Δ θ / 2) = 5611 km. Thus, v av = 5611 km / 33.5 h = 167 km / h. 7. (a) From E q. 4.11, 0 =100 + 20t – 4.9t 2 , thus t = 7 s. (b) From E q. 4.12, 0 = 400 – 2gH, so H = 120 m (c) R = v x t = 105 m. (d) v y = 20 – g t = –48.6 m / s, so v = 15i – 48.6 j m / s 8. (a) From E q. 4.11: 0 = 40 – (v sin 37°) 2 – 4.9(2 2 ), so v =16.9 m/s. Then, R = (v cos 37°)t = 27.0 m; (b) v y = –v sin 37° – 9.8t = –29.8 m / s, v x = 13.5 m / s. tan θ = v y / v x = 2.21, thus θ = 65.6° below the horizontal 9. From Example 4.2, R = v 2 sin 2 θ / g = 20.3 m From E q. 4.11: 0 = 0 + (v cos 45°)t – 4.9t 2 , so t = 2.03 s. Person must run at v = (30 m – 20.3 m) / 2.03 s = 4.77 m / s toward the thrower. 10. From Example 4.2 : sin 2 θ = R g / v 2 thus θ = 31.4° or 58.6° CHAPTER 4 11. (a) t = (18.3 m) / (30 m / s) = 0.61 s, so Δ y = –4.9 t 2 = –1.82 m (b) x = 18.3 = 30 cos θ t ; y = 0 = 30 sin θ t – 4.9t 2 . Substitute t = 18.3 / (30 cos θ ) into the equation for y to find : Sin 2 θ = 9.8 (0.61) 2 /18.3 Thus θ = 5.75° or 84.2° (reject). D = 18.3 tan 5.75° = 1.84 m OR: From Example 4.2, sin 2 θ = R g / v 2 = 0.499, so θ = 5.75° 12. (a) From x = 3 = v t and Δ y = –0.05 = –4.9t 2 , find v = 29.7 m/s; (b) From Example 4.2, sin 2 θ = R g / v 2 = 0.0333, so θ = 0.95° 13. From Example 4.2, R = v 2 sin 2 θ / g, we find v = 31.3 m / s From E q. 4.12, H = v 2 / 2g = 50 m 14. (a) 0 = 200 – v sin 37°(4) – 4.9(4 2 ), so v = 50.5 m / s ; (b) x = v cos 37° t = 161 m 15. (a) From v 2 = R g / sin 2 θ , we find v = 9.81 m / s. (b) 10.8 m/s. 16. x = v t, Δ y = –10 - 4 m = –4.9t 2 , so x = v (10 - 4 / 4.9) 1/2 = 13.6 km 17. x = 4 = v cos θ t; y = 0.8 = v sin θ t – 4.9t 2 . Substitute t = 4 / (v cos θ ) into the second equation to find v = 7 m/s....
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch04 - Exercises 1. (a) v = (6t – 2) i – 3t 2 j =...

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