Practice 3B: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Go up
Resolving vectors - by Steven Gaskill, 2001
1. How fast must a truck travel to stay beneath an airplane that is moving 105 km/h at an angle of 25 deg to the ground?
Here's what you know, va = 105 km/h, and q = 25 deg. Use the formula cos q= vx / va and plug in: cos (25 deg) = vx / 105 km/h, so vx = (105 km/h)(cos (25 deg)) = 95.1623176388 km/h = 95 km/h
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2. What is the magnitude of the vertical component of the velocity of the plane in item 1?
Here's what you know, va = 105 km/h, and q = 25 deg. Use the formula sin q= vy / va and plug in: sin (25 deg) = vy / 105 km/h, so vy = (105 km/h)(sin (25 deg)) = 44.3749174828 km/h = 44 km/h
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3. A truck drives up a hill with a 15 deg incline. If the truck has a constant speed of 22m/s, what are the horizontal and vertical components of the truck's velocity?
Here's what you know, vt = 22 m/s, and q = 15 deg. Use the formula cos q= vx / vt and plug in: cos (15 deg) = vx / 22 m/s, so vx = (22 m/s)(cos (15 deg)) = 21.2503681784 m/s = 21 m/s. Use the formula sin q = vy / vt and plug in: sin (15 deg) = vy / 22 m/s, so vy = (22 m/s)(sin (15 deg)) = 5.69401899226 m/s = 5.7 m/s. Yielding vx = 22 m/s and vy = 5.7 m/s.
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4. What are the horizontal and vertical components of a cat's displacement when it has climbed 5 m directly up a tree?
Since the tree is directly vertical the entire magnitude of 5 m is the vertical component leaving 0 m for the horizontal component.
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5. Find the horizontal and vertical components of the 125 m displacement of a superhero who flies down the top of a tall building at an angle of 25 deg below the horizontal?
Here's what you know, d = 125 m, and q = -25 deg (it's negative 'cause it's below the horizontal). Use the formula cos q= Dx / d and plug in: cos (-25 deg) = Dx / 125 m, so Dx = (125 m)(cos (-25 deg)) = 113.28847338 m = 110 m. Use the formula sin q= Dy / d and plug in: sin (-25 deg) = Dy / 125 m, so Dy = (125 m)(sin (-25 deg)) = -52.8272827176 m = -53 m. Yielding 110 m horizontal and -53 m vertical.
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6. A child rides a toboggan down a hill that decends at an angle of 30.5 deg to the horizontal. If the hill is 23.0 m long, what are the horizontal and vertical components of the child's displacement?
Here's what you know, d = 23.0 m, and q= -30.5 deg (it's negative 'cause it's decending from the horizontal). Use the formula cos q= Dx / d and plug in: cos (-30.5 deg) = Dx / 23.0 m, so Dx = (23.0 m)(cos (-30.5 deg)) = 19.8174706902 m = 19.8 m. Use the formula sin q= Dy / d and plug in: sin (-30.5 deg) = Dy / 23.0 m, so Dy = (23.0 m)(sin (-30.5 deg)) = -11.6733823481 m = -11.7 m. Yielding 19.8 m horizontal and -11.7 m vertical.
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7. A skier squats low and races down an 18 deg ski slope. During a 5 s interval the skiers accelerates at 2.5 m/s2. What are the horizontal (perpendicular to the direction of free-fall acceleration) and vertical components of the skier's acceleration during this time interval?
Here's what you know, a = 2.5 m/s2, and q= -18 deg (it's negative 'cause it's decending from the horizontal, also the 5 s is irrelevant). Use the formula cos q= ax / a and plug in: cos (-18 deg) = ax / 2.5 m/s2, so ax = (2.5 m/s2)(cos (-18 deg)) = 2.37764129074 m/s2 = 2.4 m/s2. Use the formula sin q= ay / a and plug in: sin (-18 deg) = ay / 2.5 m/s2, so ay = (2.5 m/s2)(sin (-18 deg)) = -0.772542485937 m/s2 = -0.77 m/s2. Yielding 2.4 m/s2 horizontal and -0.77 m/s2 vertical.
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