Distance and Displacement Worksheet 1 | PDF - Free Printable

Problem: Distance and Displacement Worksheet

The task involves calculating the distance and displacement for various scenarios. Let's solve each problem step by step.

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#### 1. Joey drives his Skidoo 7 km north, stops for lunch, and then drives 5 km east.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 7 \, \text{km (north)} + 5 \, \text{km (east)} = 12 \, \text{km}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point. We can use the Pythagorean theorem since the path forms a right triangle.
\[
\text{Displacement} = \sqrt{(7 \, \text{km})^2 + (5 \, \text{km})^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.6 \, \text{km}
\]

Answer:
\[
\boxed{12 \, \text{km}, 8.6 \, \text{km}}
\]

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#### 2. Anthony walks to the pizza place for lunch. He walks 1 km east, then 1 km south, and then 1 km east again.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 1 \, \text{km (east)} + 1 \, \text{km (south)} + 1 \, \text{km (east)} = 3 \, \text{km}
\]
- Displacement: To find the displacement, we determine the net change in position.
- Net change in the east direction: \(1 \, \text{km} + 1 \, \text{km} = 2 \, \text{km}\)
- Net change in the south direction: \(1 \, \text{km}\)
- Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(2 \, \text{km})^2 + (1 \, \text{km})^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.24 \, \text{km}
\]

Answer:
\[
\boxed{3 \, \text{km}, 2.24 \, \text{km}}
\]

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#### 3. Justin takes the boat 12 km south, then 4 km west, then 1 km north.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 12 \, \text{km (south)} + 4 \, \text{km (west)} + 1 \, \text{km (north)} = 17 \, \text{km}
\]
- Displacement: To find the displacement, we determine the net change in position.
- Net change in the south direction: \(12 \, \text{km} - 1 \, \text{km} = 11 \, \text{km}\)
- Net change in the west direction: \(4 \, \text{km}\)
- Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(11 \, \text{km})^2 + (4 \, \text{km})^2} = \sqrt{121 + 16} = \sqrt{137} \approx 11.7 \, \text{km}
\]

Answer:
\[
\boxed{17 \, \text{km}, 11.7 \, \text{km}}
\]

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#### 4. Preston goes on a camel safari in Africa. He travels 5 km north, then 3 km east, and then 1 km north again.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 5 \, \text{km (north)} + 3 \, \text{km (east)} + 1 \, \text{km (north)} = 9 \, \text{km}
\]
- Displacement: To find the displacement, we determine the net change in position.
- Net change in the north direction: \(5 \, \text{km} + 1 \, \text{km} = 6 \, \text{km}\)
- Net change in the east direction: \(3 \, \text{km}\)
- Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(6 \, \text{km})^2 + (3 \, \text{km})^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71 \, \text{km}
\]

Answer:
\[
\boxed{9 \, \text{km}, 6.71 \, \text{km}}
\]

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#### 5. Neil pogo sticks to his science class. He travels 3 m east and then 4 m north.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 3 \, \text{m (east)} + 4 \, \text{m (north)} = 7 \, \text{m}
\]
- Displacement: To find the displacement, we determine the straight-line distance from the starting point to the ending point. Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(3 \, \text{m})^2 + (4 \, \text{m})^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m}
\]

Answer:
\[
\boxed{7 \, \text{m}, 5 \, \text{m}}
\]

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#### 6. Alan rents a private jet for the weekend. He flies 400 km south to New York, then 700 km west to Chicago, and then 1200 km south to Miami.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 400 \, \text{km (south)} + 700 \, \text{km (west)} + 1200 \, \text{km (south)} = 2300 \, \text{km}
\]
- Displacement: To find the displacement, we determine the net change in position.
- Net change in the south direction: \(400 \, \text{km} + 1200 \, \text{km} = 1600 \, \text{km}\)
- Net change in the west direction: \(700 \, \text{km}\)
- Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(1600 \, \text{km})^2 + (700 \, \text{km})^2} = \sqrt{2560000 + 490000} = \sqrt{3050000} \approx 1746.4 \, \text{km}
\]

Answer:
\[
\boxed{2300 \, \text{km}, 1746.4 \, \text{km}}
\]

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#### 7. Brandon buys a new Seadoo. He goes 1.2 km north from the beach, jumps wakes for 6 km to the east, and then chases a boat 10 km north.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 1.2 \, \text{km (north)} + 6 \, \text{km (east)} + 10 \, \text{km (north)} = 17.2 \, \text{km}
\]
- Displacement: To find the displacement, we determine the net change in position.
- Net change in the north direction: \(1.2 \, \text{km} + 10 \, \text{km} = 11.2 \, \text{km}\)
- Net change in the east direction: \(6 \, \text{km}\)
- Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(11.2 \, \text{km})^2 + (6 \, \text{km})^2} = \sqrt{125.44 + 36} = \sqrt{161.44} \approx 12.7 \, \text{km}
\]

Answer:
\[
\boxed{17.2 \, \text{km}, 12.7 \, \text{km}}
\]

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#### 8. Alex goes cruising on his dirt bike. He rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south, 300 m east, and finally 100 m north.
- Distance: The total distance is the sum of all segments traveled.
\[
\text{Distance} = 700 \, \text{m (north)} + 300 \, \text{m (east)} + 400 \, \text{m (north)} + 600 \, \text{m (west)} + 1200 \, \text{m (south)} + 300 \, \text{m (east)} + 100 \, \text{m (north)} = 3600 \, \text{m}
\]
- Displacement: To find the displacement, we determine the net change in position.
- Net change in the north direction:
\[
700 \, \text{m} + 400 \, \text{m} + 100 \, \text{m} - 1200 \, \text{m} = 0 \, \text{m}
\]
- Net change in the east direction:
\[
300 \, \text{m} + 300 \, \text{m} - 600 \, \text{m} = 0 \, \text{m}
\]
- Since both net changes are zero, the displacement is:
\[
\text{Displacement} = \sqrt{(0 \, \text{m})^2 + (0 \, \text{m})^2} = 0 \, \text{m}
\]

Answer:
\[
\boxed{3600 \, \text{m}, 0 \, \text{m}}
\]

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#### 9. Stephen buys a new moped. He travels 3 km south and then 4 km east. How far does he need to go to get back to where he started?
- This is essentially asking for the displacement from the starting point after traveling 3 km south and 4 km east.
- Net change in the south direction: \(3 \, \text{km}\)
- Net change in the east direction: \(4 \, \text{km}\)
- Using the Pythagorean theorem:
\[
\text{Displacement} = \sqrt{(3 \, \text{km})^2 + (4 \, \text{km})^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{km}
\]

Answer:
\[
\boxed{5 \, \text{km}}
\]

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Final Answers:

1. \(\boxed{12 \, \text{km}, 8.6 \, \text{km}}\)
2. \(\boxed{3 \, \text{km}, 2.24 \, \text{km}}\)
3. \(\boxed{17 \, \text{km}, 11.7 \, \text{km}}\)
4. \(\boxed{9 \, \text{km}, 6.71 \, \text{km}}\)
5. \(\boxed{7 \, \text{m}, 5 \, \text{m}}\)
6. \(\boxed{2300 \, \text{km}, 1746.4 \, \text{km}}\)
7. \(\boxed{17.2 \, \text{km}, 12.7 \, \text{km}}\)
8. \(\boxed{3600 \, \text{m}, 0 \, \text{m}}\)
9. \(\boxed{5 \, \text{km}}\)