What is the student's displacement after walking 50m West, then 50m North, and finally 25m East?
Understand the Problem
The question is asking to calculate the student's total displacement after a series of movements in different directions: West, North, and East.
Answer
The student's displacement is approximately $55.9$ m at an angle of $63.43^\circ$ North of West.
Answer for screen readers
The student's total displacement is approximately $55.9$ m at an angle of $63.43^\circ$ North of West.
Steps to Solve
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Determine Displacement in Each Direction
The student first walks 50m West, which can be represented as a displacement vector $W = -50$ m (negative indicating the West direction).
Next, the student walks 50m North, represented as $N = +50$ m. -
Calculate Eastward Displacement
After moving north, the student moves 25m East, which is indicated as $E = +25$ m (positive indicating the East direction). -
Combine Total Displacements
The total displacement in the East-West direction can be calculated as follows: $$ \text{Total East-West Displacement} = W + E = -50 \text{ m} + 25 \text{ m} = -25 \text{ m} $$
The total displacement in the North-South direction is simply the distance moved North: $$ \text{Total North-South Displacement} = N = 50 \text{ m} $$
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Use Pythagorean Theorem for Total Displacement
Now we will find the resultant displacement using the Pythagorean theorem: $$ D = \sqrt{(\text{East-West Displacement})^2 + (\text{North-South Displacement})^2} $$
Substituting in the values: $$ D = \sqrt{(-25)^2 + (50)^2} $$ -
Calculate the Final Value
Calculate the values: $$ D = \sqrt{625 + 2500} = \sqrt{3125} $$
Now simplify: $$ D \approx 55.9 \text{ m} $$ -
Determine the Direction of Displacement
To find the angle $\theta$ of the resultant displacement relative to the West direction, use the tangent: $$ \tan(\theta) = \frac{\text{North Displacement}}{\text{West Displacement}} = \frac{50}{25} $$
Thus, $$ \theta = \tan^{-1}(2) \approx 63.43^\circ \text{ North of West} $$
The student's total displacement is approximately $55.9$ m at an angle of $63.43^\circ$ North of West.
More Information
Displacement is a vector quantity that refers to the change in position. In this case, the student's final position relative to the starting point was determined by combining the distances traveled in various directions.
Tips
- Mixing up the signs for different directions; it's crucial to know that left or down displacements usually count as negative.
- Forgetting to apply the Pythagorean theorem correctly in the final step.
