A Car’s Motion and an Observer’s Distance
A car can move quickly along a straight road while its distance from a roadside observer changes more slowly—or not at all for an instant. The key is that the observer is not on the road. The car’s position along the road and its distance from the observer are different quantities, connected by a right triangle.
We will use that triangle to find the rate at which the car’s distance from the observer changes. The direction of motion matters: the distance decreases as the car approaches the point on the road nearest the observer, then increases after the car passes it.
The Right-Triangle Model
Assume that the road is straight and that the observer remains fixed a perpendicular distance \(b\) from it, where \(b>0\). Let the point on the road nearest the observer be the reference point. Give the car a signed position \(x(t)\) measured along the road from that point. Choose the positive direction to match the car’s direction of travel. Thus, \(x<0\) before the car passes the reference point, \(x=0\) as it passes, and \(x>0\) afterward.
Let \(r(t)\) denote the car’s straight-line distance from the observer. The road segment from the reference point to the car, the perpendicular segment of length \(b\), and the segment of length \(r\) form a right triangle. The Pythagorean theorem gives the relation between the changing quantities.
The sign of \(x\) records which side of the closest point the car is on. The distance \(r\), by contrast, is always positive. Keeping those roles separate is essential when finding a signed rate of change.
Deriving the Distance Rate
Suppose \(x\) changes differentiably with time. Since \(b\) is fixed, differentiate \(r^2=x^2+b^2\) with respect to time. The chain rule applies to both changing quantities, \(r(t)\) and \(x(t)\).
Because \(b>0\), the distance \(r=\sqrt{x^2+b^2}\) is always positive, so division by \(2r\) is valid. This gives the required rate formula.
The factor \(x/r\) describes how much of the car’s motion along the road contributes to a change in its straight-line distance from the observer. Its magnitude is at most \(1\), since \(r\geq |x|\). So the magnitude of the distance rate cannot exceed the car’s speed along the road.
When the car is approaching the closest point, \(x<0\) and \(dx/dt>0\). The formula then gives \(dr/dt<0\): the distance is decreasing. After the car passes, \(x>0\) and \(dx/dt>0\), so \(dr/dt>0\): the distance is increasing. At the instant of passing, \(x=0\), and the distance rate is zero.
Worked Example: Distance Rate Before and After Passing
Worked Example: Distance Rate Before and After Passing
A straight road passes \(30\) m from a fixed observer. A car moves along the road at \(20\) m/s in the positive \(x\)-direction. Find the car’s distance rate when it is \(40\) m before the closest point, when it passes that point, and when it is \(40\) m beyond it.
Step 1: State the known values. The perpendicular distance is \(b=30\) m, and the car’s signed velocity is \(dx/dt=20\) m/s. The three positions are \(x=-40\) m, \(x=0\) m, and \(x=40\) m.
Step 2: Find the observer-distance at each position. At \(x=-40\) m, the Pythagorean relation gives \(r=\sqrt{(-40)^2+30^2}=\sqrt{1600+900}=\sqrt{2500}=50\) m. At \(x=0\), \(r=\sqrt{0^2+30^2}=30\) m. At \(x=40\) m, \(r=\sqrt{40^2+30^2}=\sqrt{1600+900}=50\) m.
Step 3: Apply the rate formula. Before passing, at \(x=-40\) m, \(dr/dt=(-40/50)(20)=(-4/5)(20)=-16\) m/s. At the closest point, at \(x=0\), \(dr/dt=(0/30)(20)=0\) m/s. After passing, at \(x=40\) m, \(dr/dt=(40/50)(20)=(4/5)(20)=16\) m/s.
Check: Substituting each position into \(r^2=x^2+b^2\) gives \(50^2=40^2+30^2\) before and after the pass, and \(30^2=0^2+30^2\) at the pass. The derived rate relation \(2r(dr/dt)=2x(dx/dt)\) also holds: before passing, \(2(50)(-16)=2(-40)(20)=-1600\); at the pass, \(2(30)(0)=2(0)(20)=0\); after passing, \(2(50)(16)=2(40)(20)=1600\). The negative, zero, and positive signs agree with the car approaching, passing, and moving away.
Why the Rate Is Zero at the Pass
At the closest point, the car is still moving along the road, so \(dx/dt\) need not be zero. What vanishes is \(x\), the horizontal leg of the right triangle. The distance rate formula therefore gives \(dr/dt=0\) at that instant.
This result also agrees with the geometry: the observer-distance is \(r=\sqrt{x^2+b^2}\), which has its minimum value \(b\) at \(x=0\). Before that point, moving toward \(x=0\) shortens the hypotenuse; after it, moving away lengthens the hypotenuse. The car’s road speed does not suddenly stop or reverse when it passes. Only the direction in which its distance from the observer is changing reverses.
Worked Example: Find the Position from a Distance Rate
Worked Example: Find the Position from a Distance Rate
An observer is \(15\) m from a straight road. A car approaches the closest point at \(20\) m/s. At what signed position is the car’s distance from the observer decreasing at \(12\) m/s?
Step 1: Use the signs and the rate formula. The car approaches the closest point, so \(x<0\). Its velocity in the chosen positive direction is \(dx/dt=20\) m/s, while the distance is decreasing at \(dr/dt=-12\) m/s. Substitute these values into the formula:
Step 2: Use the right-triangle relation to determine \(x\). Since \(r^2=x^2+15^2=x^2+225\), square \(x/r=-3/5\) and substitute for \(r^2\):
Because the car is approaching, \(x<0\). Thus \(x=-11.25\) m. The corresponding observer-distance is \(r=|x|/(3/5)=11.25(5/3)=18.75\) m.
Check: The right triangle satisfies \(18.75^2=11.25^2+15^2\), since \(351.5625=126.5625+225\). Substituting into the rate formula gives \(dr/dt=(-11.25/18.75)(20)=(-3/5)(20)=-12\) m/s, as required.
Worked Example: Distance Increasing After the Pass
Worked Example: Distance Increasing After the Pass
A car is \(15\) m beyond the closest point on a straight road. The observer is \(20\) m from the road, and the car is moving farther along the positive direction at \(12\) m/s. How quickly is its distance from the observer changing?
Step 1: Find the current distance. Here \(x=15\) m and \(b=20\) m, so \(r=\sqrt{15^2+20^2}=\sqrt{225+400}=\sqrt{625}=25\) m.
Step 2: Substitute the signed velocity. The car moves in the positive direction, so \(dx/dt=12\) m/s. Therefore, \(dr/dt=(15/25)(12)=(3/5)(12)=36/5=7.2\) m/s.
Check: The distance relation gives \(25^2=15^2+20^2\), or \(625=225+400\). The differentiated relation gives \(2r(dr/dt)=2(25)(7.2)=360\) m²/s, while \(2x(dx/dt)=2(15)(12)=360\) m²/s. The distance is increasing at \(7.2\) m/s.
A Common Mistake: Confusing Speed with Distance Rate
A car’s speed along the road is not generally the same as the rate at which its distance from an observer changes. For instance, in the first example, a tempting but incorrect calculation at \(x=-40\) m is:
This ignores that the car’s motion is not directed straight toward the observer. The car is \(40\) m along the road from the closest point, but its straight-line distance from the observer is \(50\) m. The correct calculation includes the factor \(x/r=-40/50\):
The negative sign says that the distance is decreasing. The magnitude, \(16\) m/s, is less than the car’s road speed of \(20\) m/s because only part of the car’s motion points in the direction that shortens the observer-distance.
Another error is to use only the unsigned distance along the road in place of \(x\). Doing so loses the information that the car is approaching or receding. Use a signed coordinate for position, keep \(r\) as a positive distance, and use the sign of \(dx/dt\) to represent the direction of travel.
When This Model Applies
The relation \(r^2=x^2+b^2\) assumes a fixed observer, a straight road, and a constant perpendicular separation \(b\). It applies at each instant when the car’s position along that road is represented by a differentiable function \(x(t)\). The rate formula uses the car’s signed velocity \(dx/dt\); if the car reverses direction, that velocity’s sign must change accordingly.
The model is useful because it turns a geometric picture into a precise relationship between measurable quantities. Once the geometry is expressed as an equation, differentiation connects the car’s motion along the road to the observer’s changing distance. The resulting sign also provides a direct interpretation: negative means approaching, zero means momentarily neither closer nor farther, and positive means receding.
Check Your Understanding
Use \(r^2=x^2+b^2\) and \(dr/dt=(x/r)(dx/dt)\) where appropriate.
- An observer is \(8\) m from a road. A car is at \(x=-6\) m and moves in the positive direction at \(10\) m/s. What is \(r\), and what is \(dr/dt\)?
- Why does \(dr/dt=0\) when the car is at \(x=0\), even if \(dx/dt\neq 0\)?
- A car is moving in the positive direction at \(15\) m/s while its observer-distance is decreasing. What must be the sign of \(x\)?
- For a fixed \(b>0\), explain why the magnitude of the distance rate cannot exceed the magnitude of the car’s road velocity.
- If \(b=12\) m, \(x=9\) m, and \(dx/dt=8\) m/s, find \(r\) and \(dr/dt\), including units.
