Equations of Motion with Constant Acceleration
Interactive Visualization of Kinematic Equations
Kinematic Equations (Constant Acceleration)
Example
Question:
Obtain equations of motion for constant acceleration using the method of calculus.
Solution:
First Equation:
By definition,
\[
a = \frac{dv}{dt}
\]
\[
dv = a\,dt
\]
Integrating both sides,
\[
\int_{v_0}^{v} dv = \int_{t_0}^{t} a\,dt
\]
Since \( a \) is constant,
\[
v - v_0 = a(t - t_0)
\]
If \( t_0 = 0 \), then
\[
v = v_0 + at
\]
Second Equation:
Also, \( v = \frac{dx}{dt} \), so
\[
dx = v\,dt
\]
Integrate both sides:
\[
\int_{x_0}^{x} dx = \int_{t_0}^{t} v\,dt
\]
Substitute for \( v \) from above:
\[
\int_{x_0}^{x} dx = \int_{t_0}^{t} (v_0 + at) dt
\]
\[
x - x_0 = v_0 t + \frac{1}{2} a t^2
\]
So,
\[
x = x_0 + v_0 t + \frac{1}{2} a t^2
\]
Third Equation:
We can write
\[
a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}
\]
So,
\[
v\,dv = a\,dx
\]
Integrate both sides:
\[
\int_{v_0}^{v} v\,dv = \int_{x_0}^{x} a\,dx
\]
\[
\frac{v^2 - v_0^2}{2} = a(x - x_0)
\]
\[
v^2 = v_0^2 + 2a(x - x_0)
\]
These equations hold for motion with constant acceleration. The calculus method can also be used for non-uniform acceleration.
Current Motion State
Time: 0.0 s
Velocity: 5.0 m/s
Position: 0.0 m
Simulation Parameters
Initial Velocity (v₀): 5.0 m/s
Acceleration (a): 2.0 m/s²
Initial Position (x₀): 0.0 m
Derivation of First Equation (v = v₀ + at)
By definition of acceleration:
Separate variables for integration:
Integrate both sides (from v₀ to v and 0 to t):
Since acceleration is constant:
Final velocity equation:



