Dynamics

Particle Kinematics

Constant-acceleration motion. Rectilinear: give ANY 3 of {initial velocity, final velocity, acceleration, time, displacement} and it solves the rest, with a consistency check when over-specified. Projectile: launch speed + angle (+ optional launch height) gives time of flight, range, apex height, and impact velocity. Use for: braking/accelerating vehicles, free fall, thrown or launched objects. Not for: variable acceleration or air resistance.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

moderectilinear = straight-line constant acceleration (give any 3 of the 5 motion quantities); projectile = launch at an angle under gravityinitial velocity (optional)Initial velocity v0 (rectilinear; positive along the direction of motion) [dimension: velocity]final velocity (optional)Final velocity v (rectilinear; same sign convention as v0) [dimension: velocity]acceleration (optional)Acceleration a (rectilinear; NEGATIVE for deceleration/braking) [dimension: acceleration]time (optional)Elapsed time t (rectilinear) [dimension: time]displacement (optional)Displacement s (rectilinear; along the direction of motion) [dimension: length]launch speed (optional)Launch speed (projectile mode) [dimension: velocity]launch angle deg (optional)Launch angle above horizontal in degrees (projectile mode)launch height (optional)Launch height above the landing level; omit for level ground (projectile mode) [dimension: length]

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

rectilinear from rest: a=3 m/s², t=4 s

Given

mode: rectilinear
initial velocity: 0 m/s
acceleration: 3 m/s^2
time: 4 s

Assumptions

Constant acceleration along a straight line; positive direction = direction of initial motion.

Solution steps

Choose the kinematic identities
With constant acceleration, the five motion quantities are linked by v = v0 + at, s = v0t + ½at², and v² = v0² + 2as; the given three determine the rest.
v = v0 + a·t; s = v0·t + a·t²/2; v² = v0² + 2·a·s
v0 = 0 m/s; v = 12 m/s; a = 3 m/s^2; t = 4 s; s = 24 m

Results

Initial velocity v0

0m/s

Final velocity v

12m/s

Acceleration a

3m/s^2

Elapsed time t

Displacement s

24m

Where this answer was checked

source: v = v0 + at; s = v0t + at²/2 (NCEES FE Reference Handbook kinematics) — start from rest, constant acceleration
verified by: hand-recomputed
derivation: v = 0 + 3·4 = 12 m/s; s = 0 + 3·16/2 = 24 m.

Example 2

US braking: v0=88 ft/s, v=0, a=−11.2 ft/s²

Given

mode: rectilinear
initial velocity: 88 ft/s
final velocity: 0 ft/s
acceleration: -11.2 ft/s^2

Assumptions

Constant acceleration along a straight line; positive direction = direction of initial motion.

Solution steps

Choose the kinematic identities
With constant acceleration, the five motion quantities are linked by v = v0 + at, s = v0t + ½at², and v² = v0² + 2as; the given three determine the rest.
v = v0 + a·t; s = v0·t + a·t²/2; v² = v0² + 2·a·s
v0 = 88 ft/s; v = 0 ft/s; a = -11.2 ft/s^2; t = 7.857 s; s = 345.7 ft

Results

Initial velocity v0

88ft/s

Final velocity v

0ft/s

Acceleration a

-11.2ft/s^2

Elapsed time t

7.857s

Displacement s

345.7ft

Where this answer was checked

source: v² = v0² + 2as with the AASHTO design deceleration — braking distance at 60 mph (88 ft/s)
verified by: hand-recomputed
derivation: s = (0 − 88²)/(2·(−11.2)) = 7744/22.4 = 345.71 ft; t = (0−88)/(−11.2) = 7.857 s. Matches the SSD solver's braking-distance term for 60 mph.

Example 3

projectile at 45°: v0=20 m/s, level ground

Given

mode: projectile
launch speed: 20 m/s
launch angle deg: 45

Assumptions

No air resistance; level landing surface; g = 9.80665 m/s² exactly.

Solution steps

Resolve the launch velocity
Split the launch speed into horizontal and vertical components.
v0x = v0·cosθ; v0y = v0·sinθ
v0x = 14.14 m/s; v0y = 14.14 m/s (θ = 45°)
Time of flight
Vertical motion returns to the landing level: solve y(t) = 0.
tf = (v0y + √(v0y² + 2·g·h0))/g
tf = 2.884 s
= 2.884 s
Range and apex
Horizontal velocity is constant; the apex is where vertical velocity vanishes.
R = v0x·tf; ymax = h0 + v0y²/(2g)
R = 40.79 m; apex = 10.2 m; impact speed = 20 m/s at 45° below horizontal
= 40.79 m

Results

Time of flight

2.884s

Horizontal range

40.79m

Maximum height

10.2m

Impact speed

20m/s

Impact angle below horizontal (degrees)

Where this answer was checked

source: R = v0²·sin(2θ)/g; apex = (v0 sinθ)²/2g (NCEES FE Reference Handbook) — level-ground projectile, maximum-range angle
verified by: hand-recomputed
derivation: v0x = v0y = 20/√2 = 14.1421 m/s. tf = 2·14.1421/9.80665 = 2.88424 s. R = v0²/g = 400/9.80665 = 40.7887 m. apex = 200/(2·9.80665)·... = 14.1421²/(2·9.80665) = 10.1972 m. Impact speed = launch speed = 20 m/s (energy symmetry).