Solve constant-acceleration motion or projectile flight. Find missing velocities, distance, time, range, or peak height with worked steps.
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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
4s
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)
45
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).