Distance-Time Law with Acceleration

Calculator and formulas for calculating the distance-time law with uniform acceleration

Distance-Time Law Calculator

Uniformly Accelerated Motion

Calculates the relationship between distance (s), velocities (v₀, v), acceleration (a) and time (t).

m
m/s
m/s
m/s²
s
Result

Example 2: Calculate time
Problem:

A train starts from rest (v₀ = 0) and accelerates at 2 m/s². How long does it take to travel 400 m?

Given:
  • Initial velocity: v₀ = 0 m/s
  • Acceleration: a = 2 m/s²
  • Distance: s = 400 m
  • Find: Time t
Solution:
\[s = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
\[400 = 0 + \frac{1}{2} \cdot 2 \cdot t^2\]
\[400 = t^2\]
\[t = 20 \text{ s}\]

Example Calculations

Example 1: Calculate distance
Problem:

A car drives with an initial velocity of 10 m/s and accelerates at 3 m/s² for 8 seconds. How far does it travel?

Given:
  • Initial velocity: v₀ = 10 m/s
  • Acceleration: a = 3 m/s²
  • Time: t = 8 s
  • Find: Distance s
Solution:

Distance-time law:

\[s = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
\[s = 10 \cdot 8 + \frac{1}{2} \cdot 3 \cdot 8^2\]
\[s = 80 + \frac{1}{2} \cdot 3 \cdot 64\]
\[s = 80 + 96 = 176 \text{ m}\]
The car travels 176 meters.
Example 3: Calculate acceleration
Problem:

A skateboard starts from rest and travels 50 meters in 5 seconds. What is the acceleration?

Given:
  • Initial velocity: v₀ = 0 m/s
  • Distance: s = 50 m
  • Time: t = 5 s
  • Find: Acceleration a
Solution:

Rearranging the distance-time law for a:

\[s = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
\[50 = 0 + \frac{1}{2} \cdot a \cdot 5^2\]
\[50 = \frac{1}{2} \cdot a \cdot 25\]
\[50 = 12.5 \cdot a\]
\[a = \frac{50}{12.5} = 4 \text{ m/s}^2\]

Formulas of the Distance-Time Law

The distance-time law with uniform acceleration connects all kinematic quantities. It is one of the most fundamental equations of classical mechanics.

Calculate distance

The distance-time law in its classical form.

\[s = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
s = distance [m]
v₀ = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
Calculate final velocity

Velocity after time t.

\[v = v_0 + a \cdot t\]
Linear velocity-time relationship.
Alternative: Without time

Calculation when time is unknown.

\[v^2 = v_0^2 + 2 \cdot a \cdot s\]
Important for braking distance calculations.
Calculate time

Quadratic equation from the distance-time law.

\[t = \frac{-v_0 + \sqrt{v_0^2 + 2as}}{a}\]
Positive solution of the quadratic equation.
Acceleration (Variant 1)

Calculation from velocities and time.

\[a = \frac{v - v_0}{t}\]
When v, v₀ and t are known.
Acceleration (Variant 2)

Calculation from distance and time (from rest).

\[a = \frac{2 \cdot s}{t^2}\]
When v₀ = 0, s and t are known.
Special cases
From rest (v₀ = 0):
s = ½at²
v = at
Uniform motion (a = 0):
s = v₀t
v = v₀ (constant)
Free fall (a = g ≈ 9.81 m/s²):
s = ½gt²
v = gt

Detailed Description of the Distance-Time Law

Physical Fundamentals

The distance-time law describes the relationship between the distance traveled and the elapsed time under constant acceleration. It is quadratic in time, meaning the distance grows faster with time than in uniform motion.

The equation \[s = v_0 \cdot t + \frac{1}{2} \cdot a \cdot t^2\] consists of:

  • v₀·t: The distance traveled at constant initial velocity
  • ½·a·t²: The additional distance due to acceleration
Usage Instructions

Select with the radio buttons which quantity should be calculated. Enter all known values (at least two values should be given).

Application Areas

Automotive Engineering

Acceleration distances, braking distance calculations, driving dynamics simulation. Foundation for ABS and ESP systems.

Traffic Safety

Safety distances, braking distance calculations, accident reconstruction. Critical for traffic regulations and speed limits.

Sports and Biomechanics

Sprint analysis, jump force measurement, performance optimization. Training planning and competition preparation.

Aerospace and Ballistics

Rocket launches, projectile trajectories, orbital calculations. Precise prediction of trajectories.

Graphical Representation and Interpretation
Distance-Time Diagram (s-t)

Curve shape: Parabola (quadratic)
Meaning of slope: Instantaneous velocity
Curvature: Shows the acceleration
Stronger curvature = higher acceleration

Velocity-Time Diagram (v-t)

Curve shape: Straight line (linear)
Slope: Acceleration a
y-intercept: Initial velocity v₀
Area: Corresponds to distance traveled

Practical Example: Braking Distance for Cars

A car travels at 100 km/h (27.8 m/s) and must brake suddenly. Maximum braking deceleration is approximately 8 m/s² (on dry asphalt).

Calculation of braking distance:
v₀ = 27.8 m/s, v = 0, a = -8 m/s²

\[v^2 = v_0^2 + 2as\] \[0 = 27.8^2 - 2 \cdot 8 \cdot s\] \[s = \frac{27.8^2}{16} = 48.3 \text{ m}\]

Practical significance: The braking distance is about 48 meters! This is why there are speed limits in residential areas (30 km/h braking distance ≈ 3.5 m).


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