Choose a positive direction. Displacement gives position relative to a fixed origin; velocity gives direction and speed. v= ds dt a= dv dt= d2s dt2 v= a, dt s= v, dt Find each integration constant from the given initial conditions. Do not assume the particle starts at the origin or from rest. **Quantity** | **Positive** | **Zero** | **Negative** s | Positive side of origin | At the origin | Negative side of origin v | Moves in positive direction | Instantaneously at rest | Moves in negative direction a | Velocity increasing | Velocity stationary | Velocity decreasing Speed and direction Speed=|v| va>0: speeding up va<0: slowing down A direction change requires a change in the sign of velocity. Zero velocity alone is insufficient; zero acceleration alone does not establish constant velocity. At an interior extremum of differentiable velocity, a = 0; test its nature.

Choose a positive direction. Displacement gives position relative to a fixed origin; velocity gives direction and speed.

\(v=\dfrac{\mathrm ds}{\mathrm dt}\qquad a=\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{\mathrm d^2s}{\mathrm dt^2}\)

\(v=\int a\,\mathrm dt\qquad s=\int v\,\mathrm dt\)

Find each integration constant from the given initial conditions. Do not assume the particle starts at the origin or from rest.

\(\text{Speed}=|v|\qquad va>0:\ \text{speeding up}\qquad va<0:\ \text{slowing down}\)

A direction change requires a change in the sign of velocity. Zero velocity alone is insufficient; zero acceleration alone does not establish constant velocity.

At an interior extremum of differentiable velocity, a = 0; test its nature.

Kinematics

Kinematics

Video Solution: