Kinematics Calculator (Uniform Motion, Acceleration, Free Fall, Projectile)

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when and where do two moving objects meet

Two bodies meeting: when and where they meet

In a meeting problem, the two bodies have the same position at the moment they meet. Just write S = S₀ + v·t for each, set them equal and solve for t. In uniform-motion mode, under Two bodies meeting, you give each one's initial position and velocity (with sign) and, if needed, the departure time.

How to set up and solve the meeting

With S_A = S₀A + vA·t and S_B = S₀B + vB·t, setting them equal gives t = (S₀B - S₀A)/(vA - vB). The sign of the velocity gives the direction: bodies approaching each other have opposite signs and their speeds add, while in a catch-up (same direction) they subtract. Two cars 300 km apart, at 80 km/h and 70 km/h in opposite directions, meet at t = 300/150 = 2 h, 160 km from A's origin. If the second leaves one hour later, write S_B = S₀B + vB·(t - 1).
Not every problem has a meeting. With equal velocities and different positions, the bodies stay the same distance apart and never meet. If t comes out negative, the meeting would have been in the past: two bodies moving apart already crossed before t = 0. The calculator flags both cases instead of giving a meaningless number. It also handles the body that has not left yet: the other can pass its position while it waits, and that counts as a meeting.
Another common question is when the bodies are a certain distance d apart. Set the position difference equal to +d and to -d. For A at 20 m/s starting at 0 and B at -10 m/s starting at 600 m, the difference is 600 - 30t; it equals 150 m at t = 15 s and -150 m at t = 25 s, so they are 150 m apart before and after crossing at t = 20 s, at position 400 m. The graph shows the two lines and the crossing point, and the time control shows the gap at each instant.

Six meetings worked out

Opposite directions

Input: A: 0 km, 80 km/h; B: 300 km, -70 km/h
Expected output: t = 2 h, posição 160 km

Closing speed of 150 km/h.

Delayed start

Input: A: 0, 60 km/h; B: 0, 90 km/h, parte em t = 1 h
Expected output: t = 3 h, posição 180 km

60t = 90(t - 1) gives t = 3.

In metres and seconds

Input: A: 0, 20 m/s; B: 600 m, -10 m/s
Expected output: t = 20 s, posição 400 m

600 - 30t = 0.

Same direction

Input: A: 0, 25 m/s; B: 100 m, 15 m/s
Expected output: t = 10 s, posição 250 m

The one behind gains 10 m every second.

150 m apart

Input: A: 0, 20 m/s; B: 600 m, -10 m/s; d = 150 m
Expected output: t = 15 s e t = 25 s

Before and after crossing.

Equal velocities

Input: A: 0, 10 m/s; B: 100 m, 10 m/s
Expected output: nunca se encontram

The gap stays at 100 m forever.

Full tool FAQ

What is the difference between uniform motion and constant acceleration?

In uniform motion the velocity does not change (zero acceleration) and position grows along a straight line on the s-t graph: S = S₀ + v·t. With constant acceleration the acceleration is non-zero, velocity changes linearly, v = v₀ + a·t, and the s-t graph is a parabola: S = S₀ + v₀·t + a·t²/2. Uniform motion is the special case with a = 0. A car at a steady 72 km/h covers 100 m in 5 s; one starting from rest with a = 2 m/s² covers 25 m in the same 5 s.

How do I convert km/h to m/s?

When should I use Torricelli's equation?

Which g should I use: 9.8, 10 or 9.80665?

How long does a free fall take and how fast does the object arrive?

How do I find the range and maximum height of an oblique launch?

Why do two different angles give the same range?

Why is average speed not the mean of the speeds?

What do two solutions or the negative time warning mean?

Does the calculator include air resistance?

Frequently asked questions

How do I choose the sign of the velocity?

Define an axis and an origin. A body moving in the direction where positions increase has positive velocity; one moving the other way, negative. If one car goes from A to B and another comes back, the second gets the negative sign.

What if one of the bodies has constant acceleration?

Then the position becomes S = S₀ + v₀·t + a·t²/2 and equating the two gives a quadratic with up to two roots. Use the constant-acceleration tab for each body and compare the positions, or solve the quadratic with the dedicated calculator.

Does the length of the vehicles matter?

In problems about trains or trucks overtaking, yes: the "full" passing ends only when the rear of one clears the rear of the other. For that, treat the distance covered as the sum or difference of the lengths, using the relative speed. The calculator treats bodies as points, with no size.

Next steps

Two bodies meeting: when and where they meet

How to set up and solve the meeting

With S_A = S₀A + vA·t and S_B = S₀B + vB·t, setting them equal gives t = (S₀B - S₀A)/(vA - vB). The sign of the velocity gives the direction: bodies approaching each other have opposite signs and their speeds add, while in a catch-up (same direction) they subtract. Two cars 300 km apart, at 80 km/h and 70 km/h in opposite directions, meet at t = 300/150 = 2 h, 160 km from A's origin. If the second leaves one hour later, write S_B = S₀B + vB·(t - 1).

Not every problem has a meeting. With equal velocities and different positions, the bodies stay the same distance apart and never meet. If t comes out negative, the meeting would have been in the past: two bodies moving apart already crossed before t = 0. The calculator flags both cases instead of giving a meaningless number. It also handles the body that has not left yet: the other can pass its position while it waits, and that counts as a meeting.

Another common question is when the bodies are a certain distance d apart. Set the position difference equal to +d and to -d. For A at 20 m/s starting at 0 and B at -10 m/s starting at 600 m, the difference is 600 - 30t; it equals 150 m at t = 15 s and -150 m at t = 25 s, so they are 150 m apart before and after crossing at t = 20 s, at position 400 m. The graph shows the two lines and the crossing point, and the time control shows the gap at each instant.

Six meetings worked out

Opposite directions

Input: A: 0 km, 80 km/h; B: 300 km, -70 km/h
Expected output: t = 2 h, posição 160 km

Closing speed of 150 km/h.

Delayed start

Input: A: 0, 60 km/h; B: 0, 90 km/h, parte em t = 1 h
Expected output: t = 3 h, posição 180 km

60t = 90(t - 1) gives t = 3.

In metres and seconds

Input: A: 0, 20 m/s; B: 600 m, -10 m/s
Expected output: t = 20 s, posição 400 m

600 - 30t = 0.

Same direction

Input: A: 0, 25 m/s; B: 100 m, 15 m/s
Expected output: t = 10 s, posição 250 m

The one behind gains 10 m every second.

150 m apart

Input: A: 0, 20 m/s; B: 600 m, -10 m/s; d = 150 m
Expected output: t = 15 s e t = 25 s

Before and after crossing.

Equal velocities

Input: A: 0, 10 m/s; B: 100 m, 10 m/s
Expected output: nunca se encontram

The gap stays at 100 m forever.

Full tool FAQ

What is the difference between uniform motion and constant acceleration?

How do I convert km/h to m/s?

When should I use Torricelli's equation?

Which g should I use: 9.8, 10 or 9.80665?

How long does a free fall take and how fast does the object arrive?

How do I find the range and maximum height of an oblique launch?

Why do two different angles give the same range?

Why is average speed not the mean of the speeds?

What do two solutions or the negative time warning mean?

Does the calculator include air resistance?

Frequently asked questions

Continue in the tool

More pages for this tool

Two bodies meeting: when and where they meet

How to set up and solve the meeting

Six meetings worked out

Opposite directions

Delayed start

In metres and seconds

Same direction

150 m apart

Equal velocities

Full tool FAQ

Frequently asked questions

How do I choose the sign of the velocity?

What if one of the bodies has constant acceleration?

Does the length of the vehicles matter?

Next steps

Problem data

Positions (optional)

Result

Unit conversion (to SI)

Step by step

Graphs

Position vs time

Velocity vs time

Acceleration vs time

Continue in the tool

More pages for this tool

Two bodies meeting: when and where they meet

How to set up and solve the meeting

Six meetings worked out

Opposite directions

Delayed start

In metres and seconds

Same direction

150 m apart

Equal velocities

Full tool FAQ

Frequently asked questions

How do I choose the sign of the velocity?

What if one of the bodies has constant acceleration?

Does the length of the vehicles matter?

Next steps

Problem data

Positions (optional)

Result

Unit conversion (to SI)

Step by step

Graphs

Position vs time

Velocity vs time

Acceleration vs time