When is the distance between two moving objects smallest?

Key takeaway: Whenever two position functions are given on the same line, the cleanest move is to look at the difference first. The distance is the absolute value of that difference, so the minimum usually comes from checking where the difference is zero or where the squared distance has a critical point.

Let the positions be

$s(t)=t^3+2, \qquad p(t)=t^2, \qquad t\ge 0.$

We want the distance between the two objects:

$D(t)=|s(t)-p(t)|=|t^3-t^2+2|.$

A neat way to handle this is to minimize the squared distance:

$డి(t)=\big(t^3-t^2+2\big)^2.$

Since the square is nonnegative and preserves the location of minima, we can work with the inside expression first.

1) Check whether the objects ever meet

Solve

$t^3-t^2+2=0.$

Test easy values: at $t=0$, the expression is $2$; at $t=1$, it is $2$; for $t\ge 0$, the cubic never drops below zero. In fact,

$t^3-t^2+2=t^2(t-1)+2>0$

for all $t\ge 0$.

So the distance is simply

$D(t)=t^3-t^2+2.$

2) Differentiate the distance

$D'(t)=3t^2-2t=t(3t-2).$

Critical points:

$t=0, \qquad t=\frac23.$

3) Test the critical points

Second derivative:

$D''(t)=6t-2.$

• At $t=0$, $D''(0)=-2$, so that is a local maximum.
• At $t=\frac23$,

$D''\left(\frac23\right)=2>0,$

so this is a local minimum.

Since the function grows for large $t$, this local minimum is the absolute minimum on $t\ge 0$.

Answer

The distance between the two objects is smallest at

$\boxed{t=\frac23}.$

At that time, the distance is

$D\left(\frac23\right)=\left(\frac{8}{27}-\frac{4}{9}+2\right)=\frac{50}{27}.$

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Pitfalls the pros know 👇 Don’t minimize $s(t)-p(t)$ without thinking about absolute value. Distance is never negative, so the sign matters. Another trap is ignoring the domain $t\ge 0$: the point $t=0$ must be checked, especially when a critical point lies at the boundary. If you use the squared distance, remember that it can introduce extra algebra but does not change where the minimum occurs.

What if the problem changes? If the question asked for the time when the objects are farthest apart on a closed interval, you would check both critical points and the endpoints. If one trajectory were shifted, say $s(t)=t^3+a$, the minimizing time could change or the objects might even meet, in which case the minimum distance would be $0$ at the collision time.

Tags: absolute value distance, critical point, minimum separation