Math Problem Statement
Solution
The problem states that a train moves at a constant speed of 61.0 km/h with the following movements:
- First: East for 41.0 minutes,
- Second: 49.0° east of due north for 23.0 minutes,
- Third: West for 51.0 minutes.
We are asked to determine:
- The magnitude of the average velocity.
- The angle (relative to east) of the average velocity.
Steps to Solve:
-
Convert time to hours:
-
Determine distances traveled in each segment:
- Distance in the first segment (eastward):
- Distance in the second segment (at 49.0° east of due north):
- Distance in the third segment (westward):
-
Resolve the second segment into components:
- Eastward component:
- Northward component:
-
Calculate the total displacement:
- Total eastward displacement:
- Total northward displacement:
-
Find the magnitude and direction of the average velocity:
- Magnitude:
- Angle relative to east:
-
Average velocity:
- Average velocity magnitude: , where
Let's calculate this now.### Results:
- The magnitude of the average velocity during the trip is approximately 8.90 km/h.
- The angle of the average velocity relative to east is approximately 64.00° north of east.
Would you like any further clarifications or additional details on these calculations?
Here are 5 related questions for further exploration:
- How would the average velocity change if the speed of the train was increased?
- How can you calculate the displacement if the angles of movement change?
- How would the total displacement change if the train moved west first and then east?
- What would the outcome be if the time spent moving in each direction was different?
- How is the average speed different from average velocity in this context?
Tip: When breaking a vector into components, always pay attention to the direction and angles involved, especially when resolving them into horizontal and vertical components.
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Math Problem Analysis
Mathematical Concepts
Vector addition
Trigonometry
Kinematics
Average velocity
Formulas
Magnitude of velocity: D = sqrt(D_x^2 + D_y^2)
Direction of velocity: θ = tan^(-1)(D_y/D_x)
Distance: d = speed * time
Theorems
Pythagorean theorem
Trigonometric functions for vector components
Suitable Grade Level
Grades 11-12 (Physics/Advanced Math)
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