Solve the following integral and find the equation of motion. S'do (1) dt = S'a dz

Answer 1

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.