Calculate acceleration, velocity, time, displacement, force, or mass using constant-acceleration kinematics and Newton’s second law. Solve motion problems with velocity and displacement equations or dynamics problems with F = ma while preserving direction, units, and physical meaning.
Formula Used
a = (v - v₀) / t
Acceleration is the rate at which an object changes its velocity. This calculator helps you solve problems in two key areas of physics: Kinematics (how things move) and Dynamics (why things move).
Kinematics deals with motion without considering forces. The "Big 4" kinematic equations connect displacement ($d$), velocity ($v$), acceleration ($a$), and time ($t$).
Dynamics explains motion through forces. Sir Isaac Newton's Second Law states that Force equals Mass times Acceleration:
This relationship allows you to calculate the force required to accelerate a car, a rocket, or any moving object.
Acceleration describes the rate at which velocity changes with time. Because velocity contains both magnitude and direction, acceleration can occur when an object speeds up, slows down, changes direction, or experiences some combination of those changes.
The SI unit of acceleration is meter per second squared, written m/s². NIST lists acceleration as a derived quantity with SI unit m/s².
This calculator separates two related but conceptually different physics problems: kinematics and dynamics.
Kinematics describes how an object moves without requiring a model of the forces that cause the motion. Under constant acceleration, initial velocity, final velocity, displacement, acceleration, and elapsed time are connected by a standard family of equations.
Dynamics addresses why velocity changes. Newton’s second law states that the net external force acting on a system equals mass multiplied by acceleration: F_net = ma.
The distinction is important. Knowing that a car changes velocity from 10 m/s to 25 m/s in 5 seconds is sufficient to calculate its average acceleration without knowing the engine force. Determining the net force required to produce that acceleration additionally requires the vehicle’s mass.
In the simplest kinematics mode, acceleration is calculated from velocity change divided by elapsed time: a = (v_f − v_i)/t. This gives average acceleration over the interval. If acceleration is constant, that average is also the constant acceleration throughout the interval.
Constant-acceleration motion also connects displacement to velocity and time. The equation Δx = v_i t + 1/2 at² is useful when elapsed time is known, while v_f² = v_i² + 2aΔx eliminates time entirely.
Another constant-acceleration relationship uses average velocity: Δx = [(v_i + v_f)/2]t. This is valid because velocity changes linearly with time when acceleration is constant.
The signs of velocity, acceleration, force, and displacement depend on the coordinate system. Negative acceleration does not automatically mean slowing down. An object moving in the negative direction with negative acceleration can be speeding up.
Dynamics mode uses net force, not automatically one individual applied force. If several forces act along the selected axis, they must first be combined algebraically according to direction before F_net = ma is applied.
The calculator therefore solves the mathematical relationships supplied by the selected mode. It does not automatically create a free-body diagram, determine which forces exist, or decide which direction should be positive.
The kinematic equations connect position, displacement, velocity, acceleration, and time when acceleration is constant. Newton’s second law connects acceleration to the net external force and mass of the system. The appropriate equation depends on which quantities are known and which variable must be solved.
a = (v_f − v_i)/t; v_f = v_i + at; Δx = v_i t + 1/2 at²; v_f² = v_i² + 2aΔx; Δx = (v_i + v_f)t/2; F_net = maA car increases its velocity from 10 m/s to 25 m/s during a 5-second interval.
Result: Average acceleration = 3 m/s².
The velocity increases by 3 meters per second during each second of the interval. If acceleration is constant, 3 m/s² is also the acceleration at every point in the modeled interval.
Positive acceleration points in the direction defined as positive by the coordinate system.
It does not automatically mean the object is speeding up unless velocity is also positive.
Negative acceleration points opposite the chosen positive direction.
Whether the object speeds up or slows down depends on the direction of velocity.
Zero acceleration means velocity is not changing at that instant or over the modeled constant-acceleration interval.
The velocity itself can still be nonzero.
Displacement is a signed change in position rather than total path length.
An object can travel a nonzero distance and still have zero net displacement.
Dynamics results use the net external force.
A nonzero individual force does not necessarily imply nonzero acceleration if other forces cancel it.
Velocity changes from 5 m/s to 17 m/s in 4 seconds.
a = (17 − 5)/4 = 3 m/s².
An object starts at 8 m/s and accelerates at 2 m/s² for 6 seconds.
v_f = 8 + 2 × 6 = 20 m/s.
An object starts from rest and accelerates at 4 m/s² for 5 seconds.
Δx = 0 × 5 + 1/2 × 4 × 25 = 50 m.
Velocity changes from 10 m/s to 30 m/s over 100 m.
Use v_f² = v_i² + 2aΔx to solve for acceleration.
A 1,200 kg vehicle accelerates at 2.5 m/s².
F_net = 1,200 × 2.5 = 3,000 N.
a_avg = Δv/Δt.
Velocity is a vector, so changes in direction contribute to acceleration even when speed remains constant.
v_i = 20 m/s and v_f = 5 m/s over 3 seconds.
a = (5 − 20)/3 = −5 m/s².
Acceleration sign refers to coordinate direction.
Speed changes according to the relative directions of velocity and acceleration.
Positive velocity with positive acceleration increases positive speed.
Negative velocity with negative acceleration increases the magnitude of negative velocity.
Let right be positive.
v_i = −10 m/s and a = −2 m/s².
After 3 seconds, v_f = −16 m/s, so speed increased from 10 to 16 m/s.
Positive velocity with negative acceleration reduces positive velocity.
If acceleration continues after velocity reaches zero, the object reverses direction.
v_i = 10 m/s, a = −2 m/s².
At t = 5 s, v = 0.
At t = 7 s, v = −4 m/s if the same acceleration continues.
v_f = v_i + at.
On a velocity-time graph, constant acceleration is represented by a straight-line slope.
Acceleration is dv/dt instantaneously and Δv/Δt over a finite interval.
For constant acceleration, the slope is constant.
Velocity rises from 4 to 16 m/s over 6 seconds.
Graph slope = (16 − 4)/6 = 2 m/s².
From v_f = v_i + at, t = (v_f − v_i)/a when a ≠ 0.
The sign and physical direction must remain consistent.
v_i = 10 m/s, v_f = 30 m/s, a = 4 m/s².
t = 20/4 = 5 seconds.
Rearrange v_f = v_i + at.
v_i = v_f − at.
v_f = 25 m/s, a = 3 m/s², t = 5 s.
v_i = 25 − 15 = 10 m/s.
Δx = v_i t + 1/2 at².
The first term represents displacement from initial velocity and the second represents the additional displacement produced by acceleration.
v_i = 5 m/s, a = 2 m/s², t = 4 s.
Δx = 5×4 + 1/2×2×16 = 36 m.
If v_i = 0, Δx = 1/2 at².
This relation is often used for simplified free-fall and laboratory problems.
From Δx = 1/2 at², a = 2Δx/t².
This simplified form is valid only when v_i = 0.
Δx = 80 m and t = 4 s from rest.
a = 160/16 = 10 m/s².
The missing v_i t term changes the displacement.
Use the full constant-acceleration equation.
v_f² = v_i² + 2aΔx.
It connects velocity, acceleration, and displacement directly.
a = (v_f² − v_i²)/(2Δx), provided Δx ≠ 0.
Signs and squared velocities must be handled separately and correctly.
v_i = 0, v_f = 60 m/s, Δx = 900 m.
a = 3600/1800 = 2 m/s².
The equation contains v² rather than v.
When solving backward for velocity, a square root can generate positive and negative mathematical branches.
v_f = ±√(v_i² + 2aΔx) mathematically.
The direction of motion determines which root is physically applicable.
v_avg = (v_i + v_f)/2.
This works because velocity changes linearly with time.
Δx = [(v_i + v_f)/2]t.
This equation does not explicitly contain acceleration.
v_i = 10 m/s, v_f = 30 m/s, t = 5 s.
Average velocity = 20 m/s.
Displacement = 100 m.
Velocity contains direction and is based on displacement.
Average speed is total distance divided by elapsed time.
An object can travel away from its starting point and return.
Its net displacement is zero even though it traveled a positive distance.
Signs encode direction along one selected axis.
Two- and three-dimensional motion requires vector components.
In Cartesian coordinates, a = (a_x, a_y, a_z).
Each component can be analyzed independently when the equations permit.
Magnitude is nonnegative.
A component such as a_x can be positive or negative.
Velocity changes whenever its direction changes.
Uniform circular motion is therefore accelerated motion despite constant speed.
A car travels around a curve at constant 20 m/s speed.
Its velocity direction changes continuously, so its acceleration is nonzero.
A scalar one-axis calculation cannot capture changing vector direction automatically.
Circular-motion equations or vector component analysis are required.
F_net = ma.
OpenStax defines the law using the net external force on the chosen system.
For positive mass, a = F_net/m.
The acceleration vector points in the direction of the net external force.
F_net = 100 N and m = 20 kg.
a = 100/20 = 5 m/s².
a is directly proportional to F_net.
Doubling net force doubles acceleration when mass is unchanged.
a is inversely proportional to mass.
Doubling mass halves acceleration when net force is unchanged.
100 N on 10 kg gives 10 m/s².
100 N on 50 kg gives 2 m/s².
Individual forces must be added vectorially.
Balanced forces produce zero net force and therefore zero acceleration in the Newtonian model.
50 N right and 50 N left produce F_net = 0.
The horizontal acceleration is zero.
Newton’s first law permits constant nonzero velocity when net force is zero.
Force is required to change velocity, not to maintain ideal inertial motion.
If net force is zero, an object moving at 10 m/s can continue at 10 m/s in an inertial model.
Acceleration is zero although velocity is not.
m = F_net/a when a is nonzero.
The result is positive for physically ordinary positive mass when force and acceleration directions are treated consistently.
F_net = 600 N and a = 3 m/s².
m = 200 kg.
Mass describes inertia.
Weight is the gravitational force on that mass.
Here g is gravitational acceleration.
Weight is measured in newtons while mass is measured in kilograms.
Using g_n = 9.80665 m/s², W = 10 × 9.80665.
Weight ≈ 98.0665 N.
NIST gives standard acceleration of free fall as 9.80665 m/s².
Actual local gravitational acceleration varies somewhat with location.
If acceleration is already known and only the unit needs changing, use the converter.
This calculator instead solves physical equations.
Angular acceleration describes change of angular velocity rather than linear velocity.
Its typical SI representation is radians per second squared.
A rate stated in meters per second cannot be combined directly with elapsed time entered in hours without conversion.
Use one coherent time unit throughout the equation.
2 minutes equals 120 seconds.
If velocity is in m/s, use 120 s rather than the raw number 2.
Kilometers per hour and meters per second represent the same dimension but different scales.
Convert before subtracting initial and final velocities.
72 km/h = 20 m/s.
This converted value can then be combined with SI acceleration and seconds.
Newton’s second law defines the dimensional relationship.
1 N accelerates 1 kg at 1 m/s².
For a = Δv/t, (m/s)/s reduces to m/s².
For F = ma, kg × m/s² produces newtons.
A claimed acceleration result in meters per second is missing one time dimension.
The correct acceleration unit must contain inverse time squared.
Average acceleration is Δv/Δt over a finite interval.
Instantaneous acceleration is the derivative dv/dt at a particular instant.
If acceleration truly remains constant, instantaneous acceleration equals the same value throughout the interval.
The average acceleration equals that constant value.
If a = a(t), velocity can be found by integrating acceleration over time.
The fixed constant-acceleration equations should not be applied blindly.
v(t) follows from the integral of a(t) with the initial condition.
Position then follows from integrating velocity.
This prevents users from treating the kinematic equations as universal motion formulas.
A small assumption badge near the mode selector would be useful.
Instantaneous velocity can equal zero while acceleration remains finite.
A vertically thrown object at its highest point is a classic example.
Vertical velocity is momentarily zero.
Ignoring air resistance, gravitational acceleration still points downward.
Velocity describes motion state.
Acceleration describes the rate at which that velocity is changing.
An object can move at constant velocity indefinitely in an ideal inertial frame.
Its acceleration remains zero.
For vertical motion, upward can be positive or downward can be positive.
Either convention works if used consistently.
If upward is positive, gravitational acceleration near Earth is negative.
A downward velocity is also negative under that convention.
Gravity is only one possible contribution to net acceleration.
The calculator should use gravitational acceleration only when the selected physical problem requires it.
They can contribute to net force, but their magnitude cannot be inferred from F = ma alone.
Coefficients, speed dependence, normal force, fluid properties, or other information may be required.
The user must identify the appropriate model and known variables.
A mathematically correct computation can still answer the wrong physical problem if the model is chosen incorrectly.
The calculator can retain high internal numerical precision.
A laboratory measurement with two significant figures does not become more accurate because the calculator displays ten decimals.
Convert units and evaluate with full internal precision.
Round the final displayed answer according to the application.
Show the equation selected, substituted values, converted SI values, final units, and coordinate sign.
This makes the solution auditable rather than presenting only a number.
Acceleration is the rate at which velocity changes with time.
Average acceleration is a = (v_f − v_i)/t when t represents the elapsed interval.
Meter per second squared, m/s².
It means velocity changes by a certain number of meters per second during each second under constant acceleration.
Subtract initial velocity from final velocity and divide by elapsed time.
For constant acceleration, use v_f = v_i + at.
Rearrange the same equation: v_i = v_f − at.
From v_f = v_i + at, use t = (v_f − v_i)/a when acceleration is nonzero.
For constant acceleration, one common formula is Δx = v_i t + 1/2 at².
If initial and final velocities are known, use a = (v_f² − v_i²)/(2Δx).
If v_i = 0, one option is a = v_f/t. If displacement and time are known, a = 2Δx/t².
Common forms are v_f = v_i + at, Δx = v_i t + 1/2 at², v_f² = v_i² + 2aΔx, and Δx = (v_i + v_f)t/2.
The standard forms used here assume constant acceleration during the interval.
Yes. The sign indicates direction relative to the chosen coordinate axis.
No. An object speeds up when velocity and acceleration point in the same direction and slows when they point in opposite directions.
Yes. At the highest point of a vertical throw, instantaneous vertical velocity is zero while gravitational acceleration remains downward.
Yes. Constant nonzero velocity corresponds to zero acceleration.
No. Speed measures how fast something moves; acceleration measures how velocity changes.
Velocity includes direction, while speed is the magnitude of velocity.
Displacement is signed change in position; distance is total path length traveled.
Newton’s second law states that net external force equals mass times acceleration: F_net = ma.
Use a = F_net/m.
Use F_net = ma.
Use m = F_net/a when acceleration is nonzero.
It uses the vector sum of external forces acting on the selected system.
One newton is the force required to accelerate one kilogram at one meter per second squared: 1 N = 1 kg·m/s².
No. Mass measures inertia; weight is the gravitational force acting on mass.
Standard gravity is defined as 9.80665 m/s².
Simple one-dimensional free-fall problems can use the constant-acceleration equations when gravitational acceleration is treated as constant and air resistance is neglected.
It depends on the coordinate system. If upward is positive, downward gravitational acceleration is negative.
Yes. Acceleration is a vector quantity.
Yes. An object moving in a circle at constant speed accelerates because its velocity direction changes.
Use the Angular Acceleration Calculator for rotational acceleration.
For dedicated unit conversion, use the Acceleration Converter.
Only after converting the velocity and time quantities into a consistent unit system.
Acceleration has dimensions of length/time², so integrating constant acceleration into position introduces a t² term.
The standard constant-acceleration equations are generally insufficient; variable acceleration requires integration or numerical methods.
The equations can be evaluated accurately from the entered values, but physical accuracy depends on whether constant acceleration, one-dimensional motion, force modeling, and measurement assumptions match the real situation.
Reviewed 2026-09-01 by Dr Akawak Ejigu, DBA.
Continue with calculators that answer nearby questions and help compare the next step.
Calculate force, mass, and acceleration using Newton
Convert speed among metres per second, kilometres per hour, miles per hour, feet per second, knots, and the speed of light.
Use this free angular acceleration for science calculations. Get instant, accurate results with clear inputs and no signup required.
Use this free angular impulse for science calculations. Get instant, accurate results with clear inputs and no signup required.
Use this free angular impulse momentum for science calculations. Get instant, accurate results with clear inputs and no signup required.
Use this free angular momentum for science calculations. Get instant, accurate results with clear inputs and no signup required.