Two Moving Objects Distance
About This MicroSim
This visualization helps you examine how the rate of distance change between two moving objects depends on both their positions and velocities. You'll see Car A moving north/south and Car B moving east/west from a common intersection.
The Setup
- Car A travels along a north-south road at speed \(v_A\) (mph)
- Car B travels along an east-west road at speed \(v_B\) (mph)
- Both cars start at the origin (the intersection) at time \(t = 0\)
The Distance Formula
If Car A has traveled \(a\) miles from the origin and Car B has traveled \(b\) miles, the distance between them is:
The Key Formula: Rate of Distance Change
The rate at which the distance changes is given by:
Where:
- \(\frac{da}{dt} = \pm v_A\) (positive if moving away from origin)
- \(\frac{db}{dt} = \pm v_B\) (positive if moving away from origin)
Delta Moment
"See how dd/dt isn't just about how fast each car is going? It's a weighted combination based on WHERE each car currently is! When both cars move away from the intersection, the distance grows faster than if one were heading back."
How to Use
- Car A Speed Slider: Set Car A's speed (20-100 mph)
- Car B Speed Slider: Set Car B's speed (20-100 mph)
- Direction Buttons: Choose N/S for Car A and E/W for Car B
- Time Slider: Manually scrub through time (0-5 hours)
- Play/Pause: Animate the motion
- Reset: Return to initial conditions
What to Observe
- The distance line (red dashed) connecting the two cars
- The velocity vectors (arrows) showing each car's direction
- The d(t) graph showing distance over time
- The observation panel showing current values of a, b, d, and dd/dt
- Whether the cars are getting closer or farther apart
Lesson Plan
Learning Objectives
After using this MicroSim, students will be able to:
- Analyze how the rate of distance change depends on both position and velocity (Bloom Level 4)
- Apply the related rates formula for distance between two moving objects
- Investigate how direction changes affect dd/dt
Prerequisite Knowledge
- Pythagorean theorem for distance
- Basic derivatives and rates of change
- Chain rule (helpful but not required)
Suggested Activities
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Symmetric Case: Set both cars to the same speed going away from the origin. Observe how dd/dt changes over time even though speeds are constant.
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Direction Investigation:
- Start with Car A going North, Car B going East (both away from origin)
- Note dd/dt is always positive
- Change Car A to South (toward origin)
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How does dd/dt change? When is it zero?
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Speed vs Position:
- Set Car A at 100 mph, Car B at 20 mph
- Observe which car's position contributes more to dd/dt
- Now swap: Car A at 20 mph, Car B at 100 mph
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How does the d(t) graph change?
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Finding the Minimum Distance: When both cars head toward and then past the intersection, at what time is the distance minimized? (Hint: dd/dt = 0)
Discussion Questions
- Why doesn't the distance simply increase at the sum of the two speeds?
- What happens to dd/dt when one car is very close to the origin?
- If Car A goes North at 60 mph and Car B goes West at 60 mph, is dd/dt the same as if Car B went East?
- Can dd/dt ever be greater than the speed of either car? Under what conditions?
Assessment Questions
- Car A travels north at 50 mph and Car B travels east at 40 mph. After 2 hours:
- What is the distance between them?
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What is dd/dt at that moment?
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If both cars travel away from the origin at equal speeds \(v\), show that \(dd/dt = v\sqrt{2}\) always.
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At \(t = 1\) hour, Car A is 60 miles north and Car B is 80 miles east. If Car A travels at 30 mph north and Car B at 40 mph west, is the distance increasing or decreasing?
Embedding
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