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, v

_{a}= 105 km/h, and q = 25 deg. Use the formula cos q= v_{x}/ v_{a}and plug in: cos (25 deg) = v_{x}/ 105 km/h, so v_{x}= (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, v

_{a}= 105 km/h, and q = 25 deg. Use the formula sin q= v_{y}/ v_{a}and plug in: sin (25 deg) = v_{y}/ 105 km/h, so v_{y}= (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, v

_{t}= 22 m/s, and q = 15 deg. Use the formula cos q= v_{x}/ v_{t}and plug in: cos (15 deg) = v_{x}/ 22 m/s, so v_{x}= (22 m/s)(cos (15 deg)) = 21.2503681784 m/s = 21 m/s. Use the formula sin q = v_{y}/ v_{t }and plug in: sin (15 deg) = v_{y}/ 22 m/s, so v_{y}= (22 m/s)(sin (15 deg)) = 5.69401899226 m/s = 5.7 m/s. Yielding v_{x}= 22 m/s and v_{y }= 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/s**^{2}**. 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/s

^{2}, 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= a_{x}/ a and plug in: cos (-18 deg) = a_{x}/ 2.5 m/s^{2}, so a_{x}= (2.5 m/s^{2})(cos (-18 deg)) = 2.37764129074 m/s^{2}= 2.4 m/s^{2}. Use the formula sin q= a_{y}/ a and plug in: sin (-18 deg) = a_{y}/ 2.5 m/s^{2}, so a_{y}= (2.5 m/s^{2})(sin (-18 deg)) = -0.772542485937 m/s^{2}= -0.77 m/s^{2}. Yielding 2.4 m/s^{2}horizontal and -0.77 m/s^{2}vertical.

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