The Central Idea

Two paths can intersect geometrically without two moving objects colliding.

A collision requires both:

  1. the objects occupy the same position;
  2. they occupy that position at the same time.

In vector form, objects \(A\) and \(B\) collide at time \(t\) when

\[ \boxed{ \mathbf{r}_A(t)=\mathbf{r}_B(t) }. \]

Path Intersection versus Motion Intersection

A path intersection is a geometric point that belongs to both trajectories.

A motion intersection or collision requires the objects to reach the same point at the same instant.

Therefore:

\[ \boxed{ \text{same place} + \text{same time} = \text{meeting or collision} }. \]

Opposing Motion on a Line

In one dimension, two objects move on the same line.

If they move toward one another, their separation decreases.

Suppose east is positive. Then an eastbound velocity is positive and a westbound velocity is negative.

Signed velocities allow the same relative-motion formulas to handle both directions.

General One-Dimensional Position Models

Under constant velocity, write

\[ x_A(t)=x_{A0}+v_At \]

and

\[ x_B(t)=x_{B0}+v_Bt. \]

A meeting occurs when

\[ \boxed{ x_A(t)=x_B(t) }. \]

Solving the Meeting Equation

Set

\[ x_{A0}+v_At = x_{B0}+v_Bt. \]

Rearranging,

\[ (v_A-v_B)t = x_{B0}-x_{A0}. \]

Therefore,

\[ \boxed{ t = \frac{x_{B0}-x_{A0}} {v_A-v_B} } \]

when \(v_A\ne v_B\).

Example: Objects Moving toward Each Other

Object \(A\) begins at

\[ x_A(0)=0 \]

and moves right at

\[ 6\frac{\text{m}}{\text{s}}. \]

Object \(B\) begins at

\[ x_B(0)=40 \]

and moves left at

\[ 4\frac{\text{m}}{\text{s}}. \]

Thus

\[ x_A(t)=6t \]

and

\[ x_B(t)=40-4t. \]

Set them equal:

\[ 6t=40-4t. \]

Therefore,

\[ 10t=40 \] \[ \boxed{t=4\text{ s}}. \]

Finding the Meeting Position

Substitute

\[ t=4 \]

into either position function:

\[ x_A(4)=6(4)=24. \]

Therefore,

\[ \boxed{x=24\text{ m}}. \]

Checking the other object:

\[ x_B(4)=40-4(4)=24. \]

Closing Speed

In direct opposing motion, the separation decreases at the sum of the speeds.

For speeds \(s_A\) and \(s_B\),

\[ \boxed{ \text{closing speed} = s_A+s_B }. \]

In the previous example,

\[ 6+4=10\frac{\text{m}}{\text{s}}. \]

Since the initial separation is 40 meters,

\[ t=\frac{40}{10}=4\text{ s}. \]

Moving Apart on Opposing Directions

If two objects move away from each other on the same line, their separation increases at the sum of their speeds.

For example, if one moves left at \(3\text{ m/s}\) and the other right at \(5\text{ m/s}\), the separation grows at

\[ \boxed{ 8\frac{\text{m}}{\text{s}} }. \]

Intersecting Paths in the Plane

In two dimensions, two geometric paths may cross.

For example, one object may move along

\[ y=x \]

while another moves along

\[ y=4-x. \]

The paths intersect where

\[ x=4-x. \]

Thus

\[ x=2, \qquad y=2. \]

The geometric intersection is

\[ \boxed{(2,2)}. \]

Geometric Intersection Alone Is Not Enough

Knowing that two paths intersect at

\[ (2,2) \]

does not tell us whether two moving objects collide there.

We must also determine the time at which each object reaches that point.

Separate Parameters for Separate Motions

To analyze geometric intersection alone, it can be useful to assign different parameters:

\[ \mathbf{r}_A(u) \]

and

\[ \mathbf{r}_B(v). \]

A geometric path intersection requires

\[ \boxed{ \mathbf{r}_A(u)=\mathbf{r}_B(v) } \]

for some possibly different parameter values \(u\) and \(v\).

Shared Time for Collision Analysis

When the parameter represents the same physical time for both objects, a collision requires

\[ \boxed{ \mathbf{r}_A(t)=\mathbf{r}_B(t) }. \]

The same value of \(t\) must satisfy every coordinate equation.

Example: Paths Cross but Objects Do Not Collide

Let

\[ \mathbf{r}_A(t) = \langle t,t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle4-t,4-t\rangle. \]

These objects move on the same diagonal line in opposite directions.

They meet when

\[ t=4-t, \]

so

\[ \boxed{t=2}. \]

Their common position is

\[ \boxed{(2,2)}. \]

Example: Same Geometric Crossing, Different Arrival Times

Suppose

\[ \mathbf{r}_A(t) = \langle t,t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle 2,4-t\rangle. \]

Object \(A\) reaches \((2,2)\) when

\[ t=2. \]

Object \(B\) reaches \((2,2)\) when

\[ 4-t=2, \]

so

\[ t=2. \]

Here the arrival times match, so a collision occurs.

Changing the Timing Can Remove the Collision

Now let

\[ \mathbf{r}_B(t) = \langle2,5-t\rangle. \]

Object \(B\) reaches \((2,2)\) when

\[ 5-t=2, \]

so

\[ t=3. \]

Object \(A\) still reaches \((2,2)\) at \(t=2\).

Therefore, the paths intersect geometrically, but the objects do not collide.

Collision as a System of Equations

If

\[ \mathbf{r}_A(t) = \langle x_A(t),y_A(t)\rangle \]

and

\[ \mathbf{r}_B(t) = \langle x_B(t),y_B(t)\rangle, \]

then a collision requires

\[ \boxed{ x_A(t)=x_B(t) } \]

and

\[ \boxed{ y_A(t)=y_B(t) }. \]

Both equations must hold for the same \(t\).

Example: No Collision because Times Conflict

Let

\[ \mathbf{r}_A(t) = \langle2t,t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle6-t,4\rangle. \]

The \(x\)-coordinates match when

\[ 2t=6-t, \]

giving

\[ t=2. \]

The \(y\)-coordinates match when

\[ t=4. \]

Since the same time does not satisfy both conditions, there is no collision.

Relative Position Provides a Compact Collision Test

Define

\[ \mathbf{r}_{A/B}(t) = \mathbf{r}_A(t)-\mathbf{r}_B(t). \]

A collision occurs when

\[ \boxed{ \mathbf{r}_{A/B}(t)=\mathbf{0} }. \]

This combines all coordinate conditions into one vector equation.

Example: Relative-Position Collision Test

Let

\[ \mathbf{r}_A(t) = \langle t,2t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle6-2t,6-t\rangle. \]

Then

\[ \mathbf{r}_{A/B}(t) = \langle3t-6,3t-6\rangle. \]

Set this equal to zero:

\[ 3t-6=0. \]

Therefore,

\[ \boxed{t=2}. \]

Both components vanish simultaneously.

The Collision Position

Substitute \(t=2\):

\[ \mathbf{r}_A(2) = \langle2,4\rangle. \]

Also,

\[ \mathbf{r}_B(2) = \langle2,4\rangle. \]

Therefore, the collision occurs at

\[ \boxed{(2,4)}. \]

Opposing Paths with Different Starting Times

Not every object begins moving at \(t=0\).

Suppose object \(B\) begins moving only after a delay \(t_0\).

Then a piecewise model may be required:

\[ x_B(t) = \begin{cases} x_{B0}, & t Timing must be incorporated before solving for an encounter.

Example: Delayed Departure

Object \(A\) starts at \(x=0\) and moves right at \(5\text{ m/s}\).

Object \(B\) starts at \(x=60\), waits 2 seconds, and then moves left at \(5\text{ m/s}\).

During the first 2 seconds, \(A\) travels

\[ 5(2)=10\text{ m}. \]

The remaining separation at \(t=2\) is

\[ 60-10=50\text{ m}. \]

After both are moving, closing speed is

\[ 5+5=10\text{ m/s}. \]

Additional time is

\[ \frac{50}{10}=5\text{ s}. \]

Therefore, measured from \(t=0\),

\[ \boxed{t=7\text{ s}}. \]

Path Crossing at an Angle

Two paths may cross at an angle rather than oppose each other along the same line.

For example,

\[ \mathbf{r}_A(t) = \langle t,0\rangle \]

describes horizontal motion, while

\[ \mathbf{r}_B(t) = \langle3,t-3\rangle \]

describes vertical motion.

The geometric paths cross at

\[ \boxed{(3,0)}. \]

Do the Objects Reach the Crossing Together?

Object \(A\) reaches \((3,0)\) when

\[ t=3. \]

Object \(B\) reaches \((3,0)\) when

\[ t-3=0, \]

so

\[ t=3. \]

Therefore,

\[ \boxed{\text{collision at }t=3}. \]

Intersection with Different Speeds

Suppose instead

\[ \mathbf{r}_B(t) = \langle3,2t-8\rangle. \]

Object \(B\) reaches the crossing point when

\[ 2t-8=0, \]

giving

\[ t=4. \]

Object \(A\) still reaches the crossing at \(t=3\).

The paths cross, but no collision occurs.

Three-Dimensional Intersections

In three dimensions, collision requires equality of all three coordinates:

\[ x_A(t)=x_B(t), \] \[ y_A(t)=y_B(t), \]

and

\[ z_A(t)=z_B(t). \]

Equivalently,

\[ \boxed{ \mathbf{r}_A(t)=\mathbf{r}_B(t) }. \]

Passing through the Same Point at Different Times

Suppose a space curve passes through point \(P\) more than once.

One object may reach \(P\) at time \(t_1\), while another reaches it at \(t_2\).

If

\[ t_1\ne t_2, \]

there is no collision even though both trajectories contain \(P\).

Intersection of Parametric Paths with Different Parameters

Suppose two paths are

\[ \mathbf{r}_A(u) = \langle u,u^2\rangle \]

and

\[ \mathbf{r}_B(v) = \langle v,2-v\rangle. \]

For geometric intersection, solve

\[ \langle u,u^2\rangle = \langle v,2-v\rangle. \]

Therefore,

\[ u=v \]

and

\[ u^2=2-u. \]

Solving the Geometric Intersection

Rearrange:

\[ u^2+u-2=0. \]

Factor:

\[ (u+2)(u-1)=0. \]

Thus

\[ u=-2 \]

or

\[ u=1. \]

The paths intersect at

\[ \boxed{(-2,4)} \]

and

\[ \boxed{(1,1)}. \]

Multiple Geometric Intersections

Two paths can intersect more than once.

Collision analysis then asks whether both objects reach any one of those shared points at the same physical time.

Geometry identifies candidate locations. Timing determines actual meetings.

Repeated Encounters

Periodic paths can create repeated encounters.

For example, two objects moving on the same circular track may meet again whenever their relative angular displacement changes by a full turn.

This idea will be developed more fully in uniform circular motion.

Closest Approach When No Collision Occurs

If two moving objects never occupy the same position at the same time, a natural question is:

“How close do they get?”

Define the separation vector

\[ \mathbf{s}(t) = \mathbf{r}_A(t)-\mathbf{r}_B(t). \]

Then separation distance is

\[ D(t)=|\mathbf{s}(t)|. \]

Using Squared Separation

It is often easier to minimize

\[ \boxed{ D^2(t) = \mathbf{s}(t)\cdot\mathbf{s}(t) }. \]

This avoids a square root.

The same time minimizes \(D(t)\) and \(D^2(t)\).

Example: Closest Approach

Let

\[ \mathbf{r}_A(t) = \langle t,0\rangle \]

and

\[ \mathbf{r}_B(t) = \langle4-t,3\rangle. \]

Then

\[ \mathbf{s}(t) = \langle2t-4,-3\rangle. \]

Therefore,

\[ D^2(t) = (2t-4)^2+9. \]

This is minimized when

\[ 2t-4=0. \]

Thus

\[ \boxed{t=2}. \]

The minimum separation is

\[ \boxed{3}. \]

Relative Velocity at Closest Approach

At a smooth interior minimum of separation,

\[ \frac{d}{dt}D^2(t)=0. \]

Since

\[ D^2=\mathbf{s}\cdot\mathbf{s}, \]

differentiating gives

\[ \frac{d}{dt}D^2 = 2\mathbf{s}\cdot\mathbf{s}'. \]

But

\[ \mathbf{s}' = \mathbf{v}_{A/B}. \]

Therefore, at an interior closest approach,

\[ \boxed{ \mathbf{s}\cdot\mathbf{v}_{A/B}=0 }. \]

Geometric Meaning of the Closest-Approach Condition

The equation

\[ \mathbf{s}\cdot\mathbf{v}_{A/B}=0 \]

means that the separation vector is perpendicular to the relative velocity vector at that instant.

Relative motion is then momentarily sideways with respect to the line joining the objects.

Collision Detection with Constant Relative Velocity

Suppose

\[ \mathbf{s}(t) = \mathbf{s}_0+\mathbf{v}_{\text{rel}}t, \]

where relative velocity is constant.

A collision occurs if there exists a physically allowed time \(t\) such that

\[ \boxed{ \mathbf{s}_0+\mathbf{v}_{\text{rel}}t=\mathbf{0} }. \]

Component Test for Constant Relative Velocity

In two dimensions,

\[ \mathbf{s}_0 = \langle s_x,s_y\rangle \]

and

\[ \mathbf{v}_{\text{rel}} = \langle v_x,v_y\rangle. \]

Collision requires

\[ s_x+v_xt=0 \]

and

\[ s_y+v_yt=0 \]

for the same time \(t\).

Example: Constant Relative Velocity with No Collision

Suppose

\[ \mathbf{s}_0 = \langle6,4\rangle \]

and

\[ \mathbf{v}_{\text{rel}} = \langle-2,-1\rangle. \]

The \(x\)-component reaches zero at

\[ 6-2t=0 \] \[ t=3. \]

The \(y\)-component reaches zero at

\[ 4-t=0 \] \[ t=4. \]

Since the times differ, no collision occurs.

Example: Constant Relative Velocity with Collision

Suppose

\[ \mathbf{s}_0 = \langle6,3\rangle \]

and

\[ \mathbf{v}_{\text{rel}} = \langle-2,-1\rangle. \]

Then

\[ 6-2t=0 \]

gives

\[ t=3, \]

and

\[ 3-t=0 \]

also gives

\[ t=3. \]

Therefore,

\[ \boxed{\text{collision at }t=3}. \]

Collision versus Near Miss

A collision requires zero separation.

A near miss occurs when the minimum separation is positive:

\[ \boxed{ D_{\min}>0 }. \]

The geometry of intersecting paths alone cannot distinguish these cases.

Opposing Motion with Acceleration

When velocities are not constant, use full position functions rather than a simple closing-speed shortcut.

For constant acceleration,

\[ x_A(t) = x_{A0}+v_{A0}t+\frac12a_At^2 \]

and

\[ x_B(t) = x_{B0}+v_{B0}t+\frac12a_Bt^2. \]

A meeting still requires

\[ \boxed{x_A(t)=x_B(t)}. \]

Relative Motion with Constant Acceleration

Subtract the two position models:

\[ x_{A/B}(t) = x_{A/B}(0) + v_{A/B}(0)t + \frac12a_{A/B}t^2, \]

where

\[ a_{A/B}=a_A-a_B. \]

Collision occurs when

\[ \boxed{x_{A/B}(t)=0}. \]

Example: Accelerating Approach

Suppose

\[ x_A(t)=2t+t^2 \]

and

\[ x_B(t)=30-3t. \]

Set them equal:

\[ 2t+t^2=30-3t. \]

Rearrange:

\[ t^2+5t-30=0. \]

Factor:

\[ (t+10)(t-3)=0. \]

The physically relevant nonnegative time is

\[ \boxed{t=3}. \]

Rejecting Nonphysical Time Solutions

Motion equations can produce negative algebraic roots.

If a model begins at \(t=0\), a negative solution may describe a time before the modeled interval.

Such a solution should not be accepted unless the problem explicitly allows negative time.

Path Intersection and Tangent Contact

Two paths can meet at a point without crossing transversely.

They may share the same tangent direction at the contact point.

Geometrically, this is a tangential intersection.

Whether moving objects collide there still depends on synchronized arrival.

Velocity at a Collision

At a collision time \(t_c\), compare

\[ \mathbf{v}_A(t_c) \]

and

\[ \mathbf{v}_B(t_c). \]

Their relative velocity is

\[ \boxed{ \mathbf{v}_{A/B}(t_c) = \mathbf{v}_A(t_c)-\mathbf{v}_B(t_c) }. \]

A nonzero relative velocity means the objects are passing through the shared position with different instantaneous motions.

Matching Position and Matching Velocity Are Different Conditions

Two objects can share the same position while having different velocities.

They can also have the same velocity while occupying different positions.

Therefore,

\[ \mathbf{r}_A=\mathbf{r}_B \]

and

\[ \mathbf{v}_A=\mathbf{v}_B \]

are separate conditions.

Common Error: Assuming Intersecting Paths Guarantee a Collision

A shared point on two trajectories gives only a possible meeting location.

The objects must reach that point at the same time.

Always check both geometry and timing.

Common Error: Using Different Times in a Collision Equation

Writing

\[ \mathbf{r}_A(t_1)=\mathbf{r}_B(t_2) \]

with

\[ t_1\ne t_2 \]

identifies a common geometric point, not a collision.

Collision requires one shared physical time.

Common Error: Solving Only One Coordinate

In two dimensions, solving

\[ x_A(t)=x_B(t) \]

is not enough.

We must also have

\[ y_A(t)=y_B(t) \]

at the same time.

Common Error: Adding Speeds without a Direction Model

Adding speeds works directly for simple head-on one-dimensional motion.

The general method is to assign signed velocities or vectors and compute relative velocity.

This prevents sign mistakes in more complicated situations.

Common Error: Ignoring Delayed Starts

If one object begins later than another, its position function must reflect that delay.

Using both formulas as though motion began at the same time changes the problem.

Common Error: Accepting Every Algebraic Root

A collision equation may produce several roots.

Each root must be checked against:

Common Error: Confusing Minimum Separation with Collision

At closest approach,

\[ \frac{dD}{dt}=0 \]

may hold.

But collision requires

\[ D=0. \]

A minimum distance can be positive.

A Reliable Procedure for Opposing One-Dimensional Motion

  1. Choose a positive direction.
  2. Assign signed initial positions.
  3. Assign signed velocities or full motion functions.
  4. Write \(x_A(t)\) and \(x_B(t)\).
  5. Set \(x_A(t)=x_B(t)\).
  6. Solve for time.
  7. Reject times outside the allowed domain.
  8. Substitute back to find the meeting position.

A Reliable Procedure for Intersecting Parametric Paths

  1. First decide whether the task is geometric intersection or collision.
  2. For geometric intersection, allow separate parameters if needed.
  3. For collision, use the same physical time in both motion functions.
  4. Set corresponding coordinates equal.
  5. Solve all coordinate equations.
  6. Require a common valid time for every coordinate.
  7. Substitute back to identify the shared position.

A Reliable Procedure for Closest Approach

  1. Form the separation vector \(\mathbf{s}(t)=\mathbf{r}_A(t)-\mathbf{r}_B(t)\).
  2. Compute \(D^2(t)=\mathbf{s}(t)\cdot\mathbf{s}(t)\).
  3. Differentiate \(D^2(t)\).
  4. Find critical times in the allowed domain.
  5. Check endpoints if the interval is bounded.
  6. Evaluate the separation distance at candidate times.
  7. Select the smallest valid value.

Example: Full Opposing-Motion Workflow

Two cyclists begin 150 meters apart.

Cyclist \(A\) starts at \(x=0\) and rides right at

\[ 8\frac{\text{m}}{\text{s}}. \]

Cyclist \(B\) starts at \(x=150\) and rides left at

\[ 7\frac{\text{m}}{\text{s}}. \]

Their positions are

\[ x_A(t)=8t \]

and

\[ x_B(t)=150-7t. \]

Set equal:

\[ 8t=150-7t. \]

Therefore,

\[ 15t=150 \] \[ \boxed{t=10\text{ s}}. \]

Meeting position:

\[ x=8(10) \] \[ \boxed{x=80\text{ m}}. \]

Example: Full Intersecting-Path Workflow

Let

\[ \mathbf{r}_A(t) = \langle t,2t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle6-2t,6-t\rangle. \]

Set the coordinates equal:

\[ t=6-2t \]

and

\[ 2t=6-t. \]

Both equations give

\[ 3t=6, \]

so

\[ \boxed{t=2}. \]

The collision position is

\[ \mathbf{r}_A(2) = \langle2,4\rangle. \]

Therefore,

\[ \boxed{(2,4)}. \]

Practice 1: Opposing Motion

Two objects begin 90 meters apart and move directly toward each other at \(4\text{ m/s}\) and \(5\text{ m/s}\). When do they meet?

Practice 2: Meeting Position

Object \(A\) starts at \(x=0\) and moves right at \(3\text{ m/s}\). Object \(B\) starts at \(x=40\) and moves left at \(2\text{ m/s}\). Find the meeting time and meeting position.

Practice 3: Geometric Intersection

Find the intersection of

\[ y=x+1 \]

and

\[ y=5-x. \]

Practice 4: Collision Test

Let

\[ \mathbf{r}_A(t) = \langle t,t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle4-t,2\rangle. \]

Do the objects collide?

Practice 5: Same Path, Different Times

Object \(A\) moves as

\[ \mathbf{r}_A(t)=\langle t,t\rangle. \]

Object \(B\) moves as

\[ \mathbf{r}_B(t)=\langle2,5-t\rangle. \]

Both paths pass through \((2,2)\). Do the objects collide there?

Practice 6: Relative-Position Collision

Suppose

\[ \mathbf{r}_{A/B}(t) = \langle4-2t,6-3t\rangle. \]

Determine whether a collision occurs.

Practice 7: Closest Approach

Let

\[ \mathbf{r}_A(t)=\langle t,0\rangle \]

and

\[ \mathbf{r}_B(t)=\langle6-t,4\rangle. \]

Find the time and distance of closest approach.

Practice 8: Accelerating Encounter

Let

\[ x_A(t)=t^2 \]

and

\[ x_B(t)=12-4t. \]

Find all nonnegative meeting times.

Practice 9: Three-Dimensional Collision Test

Let

\[ \mathbf{r}_A(t) = \langle t,2t,3t\rangle \]

and

\[ \mathbf{r}_B(t) = \langle4-t,8-2t,12-3t\rangle. \]

Determine whether they collide.

Practice 10: Identify the Error

A student finds that two trajectories both pass through \((3,5)\) and concludes that the objects collide.

Explain what additional information must be checked.

Practice Answers

Practice 1

Closing speed:

\[ 4+5=9\frac{\text{m}}{\text{s}}. \]

Therefore,

\[ t=\frac{90}{9} \] \[ \boxed{t=10\text{ s}}. \]

Practice 2

\[ x_A(t)=3t \]

and

\[ x_B(t)=40-2t. \]

Set equal:

\[ 3t=40-2t. \]

Thus

\[ 5t=40 \] \[ \boxed{t=8\text{ s}}. \]

Meeting position:

\[ x=3(8) \] \[ \boxed{x=24\text{ m}}. \]

Practice 3

\[ x+1=5-x. \] \[ 2x=4. \] \[ x=2. \]

Then

\[ y=3. \]

Therefore,

\[ \boxed{(2,3)}. \]

Practice 4

From the \(x\)-coordinates,

\[ t=4-t \]

gives

\[ t=2. \]

At \(t=2\),

\[ y_A=2 \]

and

\[ y_B=2. \]

Therefore,

\[ \boxed{\text{yes, they collide at }t=2}. \]

Practice 5

Object \(A\) reaches \((2,2)\) at

\[ t=2. \]

Object \(B\) reaches \((2,2)\) when

\[ 5-t=2, \]

so

\[ t=3. \]

Therefore,

\[ \boxed{\text{no collision occurs at }(2,2)}. \]

Practice 6

For collision, both components must be zero.

\[ 4-2t=0 \]

gives

\[ t=2. \]

Also,

\[ 6-3t=0 \]

gives

\[ t=2. \]

Therefore,

\[ \boxed{\text{collision at }t=2}. \]

Practice 7

\[ \mathbf{s}(t) = \langle2t-6,-4\rangle. \]

Thus

\[ D^2(t) = (2t-6)^2+16. \]

This is minimized when

\[ 2t-6=0, \]

so

\[ \boxed{t=3}. \]

Minimum separation:

\[ \boxed{D_{\min}=4}. \]

Practice 8

\[ t^2=12-4t. \]

Rearrange:

\[ t^2+4t-12=0. \]

Factor:

\[ (t+6)(t-2)=0. \]

The nonnegative solution is

\[ \boxed{t=2}. \]

Practice 9

Set the first coordinates equal:

\[ t=4-t \]

gives

\[ t=2. \]

At \(t=2\),

\[ 2t=4 \]

and

\[ 8-2t=4, \]

while

\[ 3t=6 \]

and

\[ 12-3t=6. \]

All coordinates agree.

Therefore,

\[ \boxed{\text{collision at }t=2}. \]

Practice 10

The student has shown only that the geometric paths share the point \((3,5)\).

To establish a collision, both objects must reach \((3,5)\) at the same physical time.

Their parameter or time values at that point must therefore be compared.

What to Remember

For a collision,

\[ \boxed{ \mathbf{r}_A(t)=\mathbf{r}_B(t) }. \]

Equivalently,

\[ \boxed{ \mathbf{r}_{A/B}(t)=\mathbf{0} }. \]

The core distinctions are:

When motion is synchronized by a shared time parameter, timing is part of the geometry of the problem.

Where We Go Next

Opposing and intersecting paths emphasize the relationship between geometry and timing. A particularly important recurring path is the circle, where an object can move continuously while its velocity direction changes at every instant.

The next lesson, Uniform Circular Motion, develops constant-speed motion around a circle, angular position, tangential velocity, period, frequency, and inward acceleration.