Question

side 2: for each problem:
i) draw a diagram (include the resultant as a dashed arrow)
ii.) label the diagram (numbers, units, direction)
iii.) solve for distance and displacement (include direction)
iv.) show the calculations on how you solved the problem.

1. a ufo flies 130000 km east and 50000 km west.
2. fire marshall bill climbs 12m up, 12m down, and 12m up a ladder for a drill.
3. slim bim drives his skidoo 70 kilometers north. he stops for lunch and then drives 50 kilometers west, then 70 kilometers north.
4. b-easy goes on a camel ride in northern africa. he travels 50 km north, then 60 km west and then 30 km south and finds a hidden pyramid.
5. pooh winnie must swim the length of a pool 4 times. one length of the pool is 50 m long. solve for distance and displacement. the pool is east and west.
6. a kickin’ kangaroo chases a man 10 m south, 12 m west, 7 m south, 20 m east, and 4 m north.

Answer:

Problem 1: UFO Flying

i) Diagram:

• Draw a horizontal line (east - west axis). Mark the starting point. Draw an arrow 130,000 km to the right (east), then an arrow 50,000 km to the left (west) from the end of the first arrow. The resultant (dashed arrow) is from the start to the end of the second arrow, pointing east.

ii) Labeling:

• East - west axis. First arrow: 130,000 km, east. Second arrow: 50,000 km, west. Resultant: direction east, length to be calculated.

iii) & iv) Calculations:

• Distance: Total path length. So, \( 130000 + 50000 = 180000 \) km.
• Displacement: Net change in position. East is positive, west is negative. So, \( 130000 - 50000 = 80000 \) km east.

Problem 2: Fire Marshall Bill

i) Diagram:

• Vertical line (up - down axis, up as north). Draw arrow 12m up, then 12m down (from end of first), then 12m up (from end of second). Resultant (dashed) from start to end of third arrow, up.

ii) Labeling:

• Vertical axis, up = north. Arrows: 12m up, 12m down, 12m up. Resultant: north direction, length to calculate.

iii) & iv) Calculations:

• Distance: \( 12 + 12 + 12 = 36 \) m.
• Displacement: Up is positive, down is negative. \( 12 - 12 + 12 = 12 \) m north.

Problem 3: Slim Bim

i) Diagram:

• North - south (vertical) and east - west (horizontal) axes. First arrow: 70 km north (up). Second: 50 km west (left) from end of first. Third: 70 km north (up) from end of second. Resultant: from start to end of third, with north and west components.

ii) Labeling:

• Axes: north (up), west (left). Arrows: 70 km north, 50 km west, 70 km north.

iii) & iv) Calculations:

• Distance: \( 70 + 50 + 70 = 190 \) km.
• Displacement: North: \( 70 + 70 = 140 \) km, West: 50 km. Using Pythagoras (but since we can also state components or net direction - magnitude). Magnitude: \( \sqrt{140^{2}+50^{2}}=\sqrt{19600 + 2500}=\sqrt{22100}\approx148.66 \) km. Direction: \( \arctan(\frac{50}{140})\approx19.65^{\circ} \) west of north. (Or state as 140 km north and 50 km west, but typically magnitude - direction is better. However, if just sum of north and west, but displacement is vector. So net north: 140, net west: 50. So displacement is \( \sqrt{140^{2}+50^{2}} \) km at \( \arctan(\frac{50}{140}) \) west of north.)

Problem 4: B - easy

i) Diagram:

• North - south (vertical) and east - west (horizontal) axes. First: 50 km north (up). Second: 60 km west (left) from end of first. Third: 30 km south (down) from end of second. Resultant: from start to end of third, with north - south and west components.

ii) Labeling:

• Axes: north (up), west (left), south (down). Arrows: 50 km north, 60 km west, 30 km south.

iii) & iv) Calculations:

• Distance: \( 50 + 60 + 30 = 140 \) km.
• Displacement: North - south: \( 50 - 30 = 20 \) km north. West: 60 km. Magnitude: \( \sqrt{20^{2}+60^{2}}=\sqrt{400 + 3600}=\sqrt{4000}\approx63.25 \) km. Direction: \( \arctan(\frac{60}{20}) = 71.57^{\circ} \) west of north.

Problem 5: Pooh Winnie

i) Diagram:

• Horizontal line (east - west). Draw 4 arrows: each 50 m, alternating east - west (since swimming length, so back and forth? Wait, "swim the length of a pool 4 times". One length is 50 m. So if he swims 4 lengths, assuming he starts at one end, swims to the other (50 m east), back (50 m west), then east (50 m), then west (50 m)? Wait, no: "swim the length 4 times" – maybe 4 one - way? Wait, displacement: if he starts at east end, swims west 50 m (length), then east 50 m, etc. Wait, the problem says "the pool is east and west". Let's assume:

Case 1: 4 times length, so 4 one - way (e.g., east to west, 4 times? No, length is 50 m, so each length is from one end to the other. So if he swims 4 lengths, say starting at west end:

• Swim 1: 50 m east (to east end)
• Swim 2: 50 m west (to west end)
• Swim 3: 50 m east (to east end)
• Swim 4: 50 m west (to west end)

Or maybe 4 times in the same direction? Wait, the problem says "swim the length of a pool 4 times". So distance is 4 * 50 = 200 m. Displacement: if he starts and ends at the same point (e.g., swims back and forth 2 times, which is 4 lengths: east - west - east - west), then displacement is 0. If he swims 4 times in one direction (e.g., east), but that's not a "length" (length is from one end to the other). So likely, he swims back and forth, so after 4 lengths (2 round trips), he is back at start.

ii) Labeling:

• East - west axis, length 50 m per length.

iii) & iv) Calculations:

• Distance: \( 4\times50 = 200 \) m.
• Displacement: 0 m (since he starts and ends at the same point, net change in position is zero).

Problem 6: Kickin' Kangaroo

i) Diagram:

• North - south (vertical) and east - west (horizontal) axes. First: 10 m south (down). Second: 12 m west (left) from end of first. Third: 7 m south (down) from end of second. Fourth: 20 m east (right) from end of third. Fifth: 4 m north (up) from end of fourth. Resultant: from start to end of fifth, with north - south and east - west components.

ii) Labeling:

• Axes: north (up), south (down), east (right), west (left). Arrows: 10 m south, 12 m west, 7 m south, 20 m east, 4 m north.

iii) & iv) Calculations:

• Distance: \( 10 + 12 + 7 + 20 + 4 = 53 \) m.
• Displacement:
• North - south: \( - 10 - 7 + 4=-13 \) m (south is negative? Wait, let's take north as positive, south as negative. So 10 m south: - 10, 7 m south: - 7, 4 m north: + 4. Total: \( - 10 - 7 + 4=-13 \) m (13 m south).
• East - west: \( - 12 + 20 = 8 \) m (west is negative, east is positive. 12 m west: - 12, 20 m east: + 20. Total: 8 m east.
• Magnitude: \( \sqrt{(-13)^{2}+8^{2}}=\sqrt{169 + 64}=\sqrt{233}\approx15.26 \) m.
• Direction: \( \arctan(\frac{8}{13})\approx31.6^{\circ} \) east of south.