what is position, path length and displacement
Answers
Position can be understood as the location of an object with respect to a reference point. It is expressed as a position vector.
Shown below is the position of a cyclist with respect to a lamp post. Cyclist starts at the lamp post and goes to Position 1, 2, 3 and 4 respectively.

Position 1, P1→ = +3 units
Position 2, P2→ = +5 units
Position 3, P3→ = -2units
Position 4, P4→ = -6units
Path Length/Distance
The total length of the path covered by the moving object. It is a scalar.
Following shows the distance covered by the same cyclist.

Path Length = 16 units = Distance
Displacement
Shortest distance between the initial and final positions. It is a vector.

Final Displacement, D→ = -6 units
It is interesting to observe the that final displacement vector is same as the final position vector. This is because,
P1→ + P2→ + P3→ + P4→
D→ = 3 + 2 - 7 - 4 = -6 units
General Formula of Displacement
Displacement between two positions starting at PA→ and ending at PB→ is defined as −
Displacement, D→ = PB→ - PA→
For Example, in the figure below,

Displacement between position 2 and position 4,
D→ = P4→ - P2→ = -6 - (+5) = -11 units
Distance and Displacement Relationship
Magnitude of Displacement may or may not be equal to Distance.
For example, in the figure below,

Between P1→ and P2→ −
Distance = 2 units
Magnitude of displacement = |+5 - (+3)| = 2 units
Between P1→ and P3→ −
Distance = 9 units
Magnitude of displacement = |-2 - (+3)| = 5
Let v₀ be the orbital (linear) velocity of a body revolving around the Earth, in a circular orbit of radius R.
Kinetic energy = 1/2 m v₀²,
Potential energy in Earth's gravitational field at distance R = - G Me m / R
where, m = mass of the body (or satellite),
Me = mass of Earth, G = Universal Gravitational Constant
The centripetal force for the body in the orbit is supplied by the gravitational force. Hence,
m v₀² / R = G Me m / R² => v₀² = G Me / R ---- equation 1
=> K.E. = 1/2 m v₀² = G Me m / 2 R = - P.E./ 2
Total energy at radius R = KE+PE = - G Me M / 2 R --- equation 2
Suppose now, this body is given an additional velocity v (perpendicular to the orbit and along the radius) such that it goes to a distance d from center of Earth. Since the total mechanical energy is conserved by the gravitational force, the energy at a distance d from the center of Earth is given by:
\begin{gathered}E=\frac{1}{2}mv_d^2-\frac{GM_em}{d}=\frac{1}{2}mv^2-\frac{GM_em}{2R}\\\\For\ d= > \infty,\ and\ v_{\infty}=nearly\ 0,\ minimum\ velocity\ v_e\ needed:\\\\ \frac{1}{2}mv_e^2=\frac{GM_em}{2R}\\\\v_e=\sqrt{\frac{GM_e}{R}}=v_0\\\end{gathered}
The escape velocity of a satellite is the velocity (along radius) required to send it away into the space, just manages to travel to infinite distance. It is equal to the orbital velocity.