Solve the following integral and find the equation of motion. S'do (1) dt = S'a dz
To solve this integral, we need to integrate both sides with respect to their respective variables:
∫ S'dt = ∫ S'a dz
Since S' represents the velocity and S'a represents the acceleration, we can rewrite the equation as:
∫ v dt = ∫ a dz
where v is the velocity and a is the acceleration.
Integrating both sides with respect to time, we get:
∫ v dt = ∫ a dz
∫ v dt = vz + C1
where C1 is the constant of integration.
Integrating both sides with respect to displacement (z), we get:
∫ a dz = ∫ dv dt dz
∫ a dz = ∫ dv dz dt
∫ a dz = v(dv/dz)
where dv/dz represents the gradient of the velocity curve.
Substituting this back into the original equation:
∫ v dt = ∫ a dz
vz + C1 = ∫ v(dv/dz) dz
vz + C1 = 1/2 v^2 + C2
where C2 is the constant of integration.
Rearranging the terms, we get:
1/2 v^2 - vz = C1 - C2
We can simplify this by setting C3 = C1 - C2, giving us:
1/2 v^2 - vz = C3
This is the equation of motion, which relates the velocity and displacement of an object undergoing constant acceleration.