How to Calculate Catch-Up Time When One Vehicle Starts Late

A bus passes a road marker at 8.0 m/s. Three seconds later, a car passes the same marker at 14.0 m/s and begins accelerating. The car is travelling faster, but the bus already has a head start. How long will the car take to catch it?

The key is to calculate how far ahead the bus is when the car starts, then work out how quickly that gap changes. You find the catch-up time by setting the vehicles’ positions equal and solving a uniform-acceleration equation.

There is one detail that catches students out: the two vehicles have different starting times. You must account for that before solving the equation.

01The catch-up method: start with the time difference

Imagine two vehicles travelling along the same straight road.

• Vehicle A starts moving at time \(t=0\).
• Vehicle B starts moving at time \(t=\tau\).
• Both travel in the positive direction.
• Their accelerations are constant during the period being considered.

We want to find the instant when they occupy the same position.

The easiest way to do this is to use two different time variables.

The relationship between the two clocks is:

\[ \boxed{t=\tau+s} \]

For example, if B starts 4 seconds after A and travels for another 6 seconds, the common clock reads 10 seconds.

This distinction is important because the two vehicles have different elapsed times, even though they are being observed at the same instant.

Step 1: Calculate where vehicle A is when B starts

We use the uniform-acceleration equation:

\[ x=x_0+ut+\frac12at^2 \]

Here:

• \(x\) is the final position, in metres.
• \(x_0\) is the initial position, in metres.
• \(u\) is the initial velocity, in m/s.
• \(a\) is the constant acceleration, in m/s².
• \(t\) is the elapsed time, in seconds.

For vehicle A, its elapsed time when B starts is \(\tau\).

Therefore:

\[ x_A(\tau)=x_{A0}+u_A\tau+ \frac12a_A\tau^2 \]

We also need A’s velocity at that instant:

\[ v_A(\tau)=u_A+a_A\tau \]

This gives us two useful pieces of information:

1. How far ahead A is when B starts.
2. How fast A is moving when B starts.

Define the initial gap as:

\[ D=x_A(\tau)-x_{B0} \]

Here, \(x_{B0}\) is B’s starting position.

If A is ahead, \(D\) is positive.

Step 2: Describe both vehicles using the same elapsed time

Once B starts, let \(s\) represent the time that has passed since its release.

Vehicle A has already been moving for \(\tau\) seconds, so its position after another \(s\) seconds is:

\[ x_A=x_A(\tau)+v_A(\tau)s+ \frac12a_As^2 \]

Vehicle B starts at \(x_{B0}\), so:

\[ x_B=x_{B0}+u_Bs+ \frac12a_Bs^2 \]

Notice that both equations now use \(s\).

This is the main advantage of resetting the time variable when B starts. We can compare both vehicles at exactly the same instant without mixing their different starting times.

Step 3: Set their positions equal

The vehicles meet when:

\[ x_A=x_B \]

Substituting the two position equations gives:

\[ x_A(\tau)+v_A(\tau)s+ \frac12a_As^2 = x_{B0}+u_Bs+\frac12a_Bs^2 \]

Rearranging:

\[ \boxed{ D+[v_A(\tau)-u_B]s+ \frac12(a_A-a_B)s^2=0 } \]

This is the general catch-up equation.

It describes the separation between the vehicles as time passes.

The three terms represent:

• \(D\): the initial gap when B starts.
• \(v_A(\tau)-u_B\): their initial relative velocity.
• \(a_A-a_B\): their relative acceleration.

The answer is found by solving this equation for \(s\).

Why relative motion makes the calculation easier

You can think of the problem as watching the gap between the vehicles rather than tracking both vehicles separately.

If B is faster than A, the gap might shrink.

If A accelerates more strongly, the gap might start shrinking but then increase again.

If B accelerates more strongly, the gap might close faster as time passes.

The relative-motion equation accounts for all three possibilities.

02Understanding the roots: which answer is physically correct?

The catch-up equation is usually a quadratic:

\[ \alpha s^2+\beta s+\gamma=0 \]

where:

\[ \alpha=\frac12(a_A-a_B) \]

\[ \beta=v_A(\tau)-u_B \]

\[ \gamma=D \]

If \(\alpha\neq0\), use the quadratic formula:

\[ s=\frac{-\beta\pm \sqrt{\beta^2-4\alpha\gamma}}{2\alpha} \]

But getting two numerical answers does not mean both are valid catch-up times.

You must interpret the roots.

There is another important condition: the acceleration must remain constant until the calculated meeting.

Suppose B is braking and comes to rest after 10 seconds. A mathematical root at 15 seconds cannot automatically be accepted. The braking equation no longer describes B’s motion after it stops.

You would need to calculate the next stage separately.

What if both vehicles have the same acceleration?

If:

\[ a_A=a_B \]

the quadratic term disappears.

The catch-up equation becomes:

\[ D+[v_A(\tau)-u_B]s=0 \]

Therefore:

\[ \boxed{ s=\frac{D}{u_B-v_A(\tau)} } \]

provided the relative velocity is nonzero.

Equal acceleration does not mean equal velocity.

If B is initially faster, it can still catch A. The gap simply changes at a constant rate.

05Testing the prediction with an experiment

The same reasoning can be tested using two dynamics carts.

Imagine two horizontal, parallel tracks. Each track has a cart connected by a light string over a pulley at the far end to a hanging mass.

The hanging masses pull the carts towards the pulleys. If the masses, frictional conditions, and total system masses remain constant, each cart can be modelled as having approximately constant acceleration over a suitable interval.

Place a ruler alongside each track and attach a visible marker to each cart. An overhead camera records both carts in the same video.

What should the experiment show?

Release cart A at \(t=0\).

After a measured delay \(\tau\), release cart B.

The camera records both positions against one shared time reference.

If the uniform-acceleration model is suitable, each cart’s position should follow a curved position-time relationship. A catch-up occurs when their longitudinal positions become equal.

The key observation is not simply that B moves faster. It is that the measured separation between the carts reaches zero.

Why using separate stopwatches is a problem

Suppose one stopwatch starts when cart A is released and another starts when cart B is released.

Their readings cannot be compared directly.

A reading of 3.0 seconds on the first stopwatch represents a different instant from 3.0 seconds on the second stopwatch.

A better design uses one camera clock or one synchronised timing system.

Then:

\[ t=\tau+s \]

converts the second cart’s elapsed time into the common time reference.

This is exactly the correction needed in the mathematical method.

Estimating the timing resolution

Suppose the camera records at 100 frames per second.

The time between consecutive frames is:

\[ \Delta t=\frac{1}{100} \]

\[ \boxed{\Delta t=0.010\ \mathrm{s}} \]

A cart travelling at 8.0 m/s moves approximately:

\[ \Delta x=v\Delta t \]

\[ \Delta x=8.0(0.010) \]

\[ \boxed{\Delta x=0.080\ \mathrm{m}} \]

That is 8 cm between frames.

The exact meeting may occur between two recorded frames, so the experiment has finite timing and position resolution. A higher frame rate or interpolation between adjacent positions can improve the estimate.

There is also an important limitation. If a hanging mass reaches the floor, the pulling force changes and the cart may no longer have the same acceleration. The constant-acceleration model should only be applied up to that point, unless the next stage is modelled separately.

The experiment therefore tests two things: whether the measured motion is approximately uniformly accelerated and whether the predicted meeting time agrees with the observed position crossing.

06Questions and solutions

The following questions build from a standard delayed-start calculation to situations involving multiple meetings, inverse timing, and changes in acceleration.

Question 1

Two motorbikes travel along parallel lanes in the same direction.

Motorbike A passes a road marker at \(t=0\), travelling at a constant 7.0 m/s.

Motorbike B passes the same marker 4.0 seconds later, travelling at a constant 11.0 m/s.

Calculate:

a) The time after A starts when B catches A.

b) The position where they meet, measured from the original marker.

Solution 1

B catches A 11.0 seconds after A passes the marker, at a position of 77 m.

First, calculate A’s lead when B starts:

\[ D=7.0(4.0) \]

\[ D=28\ \mathrm{m} \]

Both motorbikes have constant velocity, so their relative acceleration is zero.

The separation equation becomes:

\[ 28+(7.0-11.0)s=0 \]

\[ 28-4s=0 \]

Therefore:

\[ s=\frac{28}{4} \]

\[ \boxed{s=7.0\ \mathrm{s}} \]

This is the time after B starts.

The common time is:

\[ t=4.0+7.0 \]

\[ \boxed{t=11.0\ \mathrm{s}} \]

Calculate the meeting position using A:

\[ x_A=7.0(11.0) \]

\[ \boxed{x=77\ \mathrm{m}} \]

Checking with B:

\[ x_B=11.0(7.0)=77\ \mathrm{m} \]

Both positions agree.

Question 2

Vehicle A travels at a constant 10.0 m/s from a road marker at \(t=0\).

Vehicle B passes the same marker 5.0 seconds later, travelling at 20.0 m/s. It then experiences a constant acceleration of \(-0.50\ \mathrm{m\,s^{-2}}\).

Assume B continues braking uniformly until it reaches rest.

Calculate:

a) All the times after A starts when the vehicles occupy the same position.

b) Whether B is still travelling forwards at each meeting.

c) Which vehicle is travelling faster at each meeting.

Solution 2

The vehicles meet twice: at approximately 10.86 s and 39.14 s after A starts. B is still moving forwards at both meetings. It is faster at the first meeting, while A is faster at the second.

First, calculate A’s position when B starts:

\[ x_A=10.0(5.0) \]

\[ D=50\ \mathrm{m} \]

The relative-motion equation is:

\[ 50+(10-20)s+ \frac12[0-(-0.50)]s^2=0 \]

Simplifying:

\[ 50-10s+0.25s^2=0 \]

Multiply by 4:

\[ s^2-40s+200=0 \]

Using the quadratic formula:

\[ s=\frac{40\pm\sqrt{1600-800}}{2} \]

\[ s=\frac{40\pm\sqrt{800}}{2} \]

The two roots are:

\[ s_1\approx5.86\ \mathrm{s} \]

\[ s_2\approx34.14\ \mathrm{s} \]

Both roots are positive.

Convert to the common clock:

\[ t_1=5.0+5.86 \]

\[ \boxed{t_1=10.86\ \mathrm{s}} \]

And:

\[ t_2=5.0+34.14 \]

\[ \boxed{t_2=39.14\ \mathrm{s}} \]

Now check whether B is still moving forwards.

Its velocity is:

\[ v_B=u_B+a_Bs \]

At the first meeting:

\[ v_B=20.0-0.50(5.86) \]

\[ \boxed{v_B\approx17.07\ \mathrm{m\,s^{-1}}} \]

At the second meeting:

\[ v_B=20.0-0.50(34.14) \]

\[ \boxed{v_B\approx2.93\ \mathrm{m\,s^{-1}}} \]

Both velocities are positive.

B reaches rest after:

\[ 0=20.0-0.50s \]

\[ s=40.0\ \mathrm{s} \]

Both meetings occur before this time, so both roots are physically valid.

At the first meeting, B is moving faster and overtakes A.

At the second meeting, B has slowed sufficiently for A to catch it.

This is why two positive roots can represent two genuine meetings rather than a mathematical mistake.

Question 3

At \(t=3.0\) seconds, vehicle A is 24 m ahead of a road marker and travelling at 9.0 m/s.

At that instant, vehicle B passes the marker travelling at 13.0 m/s.

From this moment onwards, both vehicles experience the same constant acceleration of \(+0.60\ \mathrm{m\,s^{-2}}\).

A student claims:

“Because both vehicles have the same acceleration, the distance between them cannot change.”

Is this correct?

Calculate:

a) The time after A starts when B catches A.

b) The position where they meet.

c) Whether another meeting is possible under the stated conditions.

Solution 3

The student is incorrect. B catches A at \(t=9.0\) seconds, at position \(88.8\) m. There cannot be a second meeting while both vehicles maintain the same constant acceleration.

At B’s starting instant:

\[ D=24\ \mathrm{m} \]

The relative-motion equation is:

\[ 24+(9.0-13.0)s+ \frac12(0.60-0.60)s^2=0 \]

The acceleration terms cancel:

\[ 24-4s=0 \]

Therefore:

\[ \boxed{s=6.0\ \mathrm{s}} \]

The common time is:

\[ t=3.0+6.0 \]

\[ \boxed{t=9.0\ \mathrm{s}} \]

Calculate the meeting position using A:

\[ x_A=24+9.0(6.0)+ \frac12(0.60)(6.0)^2 \]

\[ x_A=24+54+10.8 \]

\[ \boxed{x_A=88.8\ \mathrm{m}} \]

Check using B:

\[ x_B=13.0(6.0)+ \frac12(0.60)(6.0)^2 \]

\[ x_B=78+10.8 \]

\[ \boxed{x_B=88.8\ \mathrm{m}} \]

The positions agree.

The key idea is that equal acceleration means the velocity difference remains constant. It does not mean the velocities are equal.

Here, B starts 4.0 m/s faster, so the 24 m gap closes at a constant rate of 4.0 m/s.

Because the relative velocity never changes, the separation decreases steadily until the vehicles meet. It cannot reverse direction and produce a second meeting.

Question 4

Vehicle A passes a road marker at \(t=0\), travelling at a constant 6.0 m/s.

Vehicle B passes the same marker at an unknown time \(\tau\), travelling at 15.0 m/s. B then accelerates uniformly at \(0.50\ \mathrm{m\,s^{-2}}\).

A second marker is located 150 m from the starting point.

Calculate the latest possible starting delay \(\tau\) that allows B to catch A at or before the 150 m marker.

Assume both vehicles maintain their stated motion until they reach the marker.

Solution 4

The latest possible delay is approximately 16.27 seconds. At this limiting delay, B catches A exactly at the 150 m marker.

First, calculate how long A takes to reach the second marker.

\[ t_A=\frac{x}{v} \]

\[ t_A=\frac{150}{6.0} \]

\[ \boxed{t_A=25.0\ \mathrm{s}} \]

For B to catch A at the marker at the latest possible time, B must also reach 150 m at \(t=25.0\) seconds.

Let \(s\) be the time B travels after its delayed start.

Using uniform acceleration:

\[ x_B=u_Bs+\frac12a_Bs^2 \]

Substitute:

\[ 150=15.0s+\frac12(0.50)s^2 \]

\[ 150=15s+0.25s^2 \]

Rearrange:

\[ 0.25s^2+15s-150=0 \]

Multiply by 4:

\[ s^2+60s-600=0 \]

Using the quadratic formula:

\[ s=\frac{-60\pm\sqrt{60^2-4(1)(-600)}}{2} \]

\[ s=\frac{-60\pm\sqrt{6000}}{2} \]

The positive root is:

\[ \boxed{s\approx8.73\ \mathrm{s}} \]

The negative root is physically invalid because it represents a time before B starts.

Now calculate the latest delay:

\[ t_A=\tau+s \]

\[ 25.0=\tau+8.73 \]

\[ \boxed{\tau\approx16.27\ \mathrm{s}} \]

If B starts earlier, it catches A before the 150 m marker.

If B starts later, it cannot catch A before A reaches that marker.

This is an inverse catch-up problem: instead of finding the meeting time from a known delay, we use the required meeting position to calculate the latest possible delay.

Question 5

Two vehicles travel along parallel lanes in the same direction. Their longitudinal positions are measured from the same road marker.

Vehicle A:

• Passes the marker at \(t=0\).
• Has an initial velocity of 2.0 m/s.
• Accelerates uniformly at \(+0.50\ \mathrm{m\,s^{-2}}\) for the first 8.0 seconds.
• Then continues at the constant velocity it has reached.

Vehicle B:

• Passes the same marker at \(t=4.0\) seconds.
• Has an initial velocity of 15.0 m/s.
• Brakes uniformly at \(-0.50\ \mathrm{m\,s^{-2}}\) until it reaches rest.
• Remains stationary after stopping.

Find every time at which the vehicles have the same longitudinal position.

Explain why simply solving one quadratic equation for the entire motion gives an incorrect result.

Solution 5

The vehicles meet twice: first at approximately \(t=5.15\) seconds, then again at approximately \(t=40.17\) seconds.

The first meeting occurs while both vehicles are accelerating or braking uniformly. The second occurs after B has stopped and A is travelling at constant velocity.

The important challenge is recognising that the equations describing the motion change at different times.

Step 1: Calculate A’s position and velocity when B starts.

B starts at \(t=4.0\) seconds.

A has been accelerating for 4.0 seconds.

Its position is:

\[ x_A=2.0(4.0)+ \frac12(0.50)(4.0)^2 \]

\[ \boxed{x_A=12.0\ \mathrm{m}} \]

Its velocity is:

\[ v_A=2.0+0.50(4.0) \]

\[ \boxed{v_A=4.0\ \mathrm{m\,s^{-1}}} \]

B starts at the marker with a velocity of 15.0 m/s.

Therefore, the initial separation is 12.0 m.

Step 2: Find the first possible meeting.

Let \(s\) be the time after B starts.

For the first 4.0 seconds after B starts, both vehicles have constant acceleration:

\[ a_A=+0.50 \]

\[ a_B=-0.50 \]

The relative-motion equation becomes:

\[ 12+(4-15)s+ \frac12(0.50-(-0.50))s^2=0 \]

Simplifying:

\[ 12-11s+0.50s^2=0 \]

Multiply by 2:

\[ s^2-22s+24=0 \]

Using the quadratic formula:

\[ s=\frac{22\pm\sqrt{22^2-4(24)}}{2} \]

\[ s=\frac{22\pm\sqrt{388}}{2} \]

The two roots are approximately:

\[ s_1=1.15\ \mathrm{s} \]

\[ s_2=20.85\ \mathrm{s} \]

However, A only maintains its acceleration for another 4.0 seconds.

Therefore, the root at 20.85 seconds is outside the interval in which this equation is valid.

Only the first root is physically acceptable for this stage.

The first meeting occurs at:

\[ t=4.0+1.15 \]

\[ \boxed{t_1=5.15\ \mathrm{s}} \]

Step 3: Calculate the positions and velocities when A changes to constant velocity.

A stops accelerating at the common time:

\[ t=8.0\ \mathrm{s} \]

Its velocity is:

\[ v_A=2.0+0.50(8.0) \]

\[ \boxed{v_A=6.0\ \mathrm{m\,s^{-1}}} \]

Its position is:

\[ x_A=2.0(8.0)+ \frac12(0.50)(8.0)^2 \]

\[ \boxed{x_A=32.0\ \mathrm{m}} \]

At this same instant, B has travelled for 4.0 seconds.

Its position is:

\[ x_B=15(4.0)+ \frac12(-0.50)(4.0)^2 \]

\[ \boxed{x_B=56.0\ \mathrm{m}} \]

Its velocity is:

\[ v_B=15-0.50(4.0) \]

\[ \boxed{v_B=13.0\ \mathrm{m\,s^{-1}}} \]

B is now 24 m ahead of A.

Step 4: Check whether another meeting occurs while B is still braking.

Let \(q\) be the time after \(t=8.0\) seconds.

During this interval:

• A travels at a constant 6.0 m/s.
• B continues braking at \(-0.50\ \mathrm{m\,s^{-2}}\).

The relative-motion equation becomes:

\[ -24+(6-13)q+ \frac12(0-(-0.50))q^2=0 \]

Simplifying:

\[ -24-7q+0.25q^2=0 \]

Rearrange:

\[ 0.25q^2-7q-24=0 \]

Multiply by 4:

\[ q^2-28q-96=0 \]

The positive root is:

\[ q=\frac{28+\sqrt{28^2+4(96)}}{2} \]

\[ \boxed{q\approx31.09\ \mathrm{s}} \]

But this result needs another check.

B does not brake indefinitely.

Its stopping time, measured from its own start, is:

\[ 0=15-0.50s \]

\[ s=30.0\ \mathrm{s} \]

Since B started at \(t=4.0\) seconds, it stops at:

\[ t=34.0\ \mathrm{s} \]

The interval after \(t=8.0\) seconds therefore lasts only:

\[ q=34-8=26.0\ \mathrm{s} \]

The predicted meeting at \(q=31.09\) seconds occurs after B has stopped.

It is not a valid solution for this braking stage.

Step 5: Calculate where the vehicles are when B stops.

B’s total displacement from its starting point is:

\[ x_B=15(30)+ \frac12(-0.50)(30)^2 \]

\[ x_B=450-225 \]

\[ \boxed{x_B=225\ \mathrm{m}} \]

A has travelled at 6.0 m/s for 26 seconds since \(t=8.0\) seconds.

Its position is:

\[ x_A=32+6.0(26) \]

\[ \boxed{x_A=188\ \mathrm{m}} \]

Therefore, when B stops, A is still behind by:

\[ D=225-188 \]

\[ \boxed{D=37\ \mathrm{m}} \]

Step 6: Calculate the final meeting.

After B stops, its position remains constant at 225 m.

A continues travelling at 6.0 m/s.

The time A needs to cover the remaining 37 m is:

\[ s=\frac{37}{6.0} \]

\[ \boxed{s=6.17\ \mathrm{s}} \]

This time is measured from B stopping at \(t=34.0\) seconds.

Therefore:

\[ t_2=34.0+6.17 \]

\[ \boxed{t_2=40.17\ \mathrm{s}} \]

At this instant, both vehicles occupy the same position:

\[ \boxed{x=225\ \mathrm{m}} \]

The two meetings are therefore:

The tempting mistake is to extend one quadratic equation across the entire motion. That produces roots which may be mathematically correct for the assumed equation but physically impossible for the actual vehicles.

The correct approach is to check when each vehicle changes acceleration, split the motion into intervals, and solve each interval using its own valid equations.

07What to check next

When a catch-up question involves only constant velocities or constant accelerations, the relative-motion equation gives you a direct method.

When a vehicle brakes, stops, or changes acceleration, you need to check whether your calculated meeting time falls inside the interval for which the equation is valid.

Two useful related concepts are:

• Can an Object Have Zero Velocity and Nonzero Acceleration? – particularly useful when a vehicle momentarily reaches zero velocity but is still accelerating.
• How to Calculate Acceleration from Speed and Stopping Distance – useful when you need to determine the braking acceleration or check when a vehicle reaches rest.

The final habit to develop is simple: never accept a catch-up root until you have checked both its time origin and the motion conditions that apply at that instant.