1. Displacement:

An object moves along the path from the left toward the right. In the reference frame xOy, its vector position is ⃗r₁at the time t₁, and ⃗r₂ at the time t₂. Then its displacement over the period of time Δt = t₂ - t₁ is Δ⃗r = ⃗r₂ - ⃗r₁. We can write:
⃗r₁ = x₁ ⃗i + y₁⃗j, and ⃗r₂= x₂ ⃗i + y₂⃗j
Then: Δ⃗r= [(x₂- x₁)⃗i + (y₂- y₁)⃗j

2. Velocity:

We will use three important terms relating to velocity.
The average velocity of the object is defined as the rate of the change of the displacement, that is ⃗v= Δ⃗r/Δt. Remark ⃗v and Δ⃗r are parallel and have the same direction (because they are two directly proportional vectors).

The instantaneous velocity or velocity for short is the limit of the average velocity when the measured period of time between two displacement is short. That is, if we assume t₁ fixed, and t₂ approaches t₁, Δt tends to zero. Thus, the average velocity becomes its derivative:

⃗v = lim (Δ⃗r/Δt) = d⃗r/dt
        Δt → 0

The speed is the magnitude of the velocity. In SI units, it is measured in meter per second (m/s or m s^-1).

Examples of speeds:
Our walking speed is about 1 m/s,
The speed of light is about 3 x 10⁸ ms^-1.

We can write:
⃗r = x ⃗i+ y ⃗j = d⃗r/dt = (dx /dt)⃗i+ (dy/dt)⃗j

3. Acceleration:

An acceleration is defined as the rate of the change of the velocity. That is:
⃗a = lim (Δ⃗v/Δt) = d⃗v/dt
        Δt → 0
Remark also ⃗a and Δ⃗v are parallel and have the same direction (because of their proportionality).

We have three cases depending on the related speeds (magnitude of velocities):

acceleration: when v2 > v1.

deceleration: when v2 < v1.

uniform motion: when v2 = v1.

limit case: when dt tends to 0.

In SI units, the magnitude of an acceleration is measured in meter per second (m/s/s or m/s² or m s^-2).



Example of acceleration:
The acceleration of gravity called gravitation is equal to about 9.8 m s^-2.

4. Expression of the acceleration in Cartesian coordinates:

We can write:
⃗r = x ⃗i+ y ⃗j

⃗v = v_x ⃗i+ v_y ⃗j
⃗v = (dx/dt) ⃗i+ (dy/dt) ⃗j

⃗a = (dv_x/dt) ⃗i+ (dv_y/dt) ⃗j
⃗a = (d²x/dt²) ⃗i+ (d²y/dt²) ⃗j
⃗a = (a_x/dt) ⃗i+ (a_y/dt) ⃗j
⃗a = d²⃗r/dt²

5. General motion: Expression of the acceleration in polar coordinates system:

At the time "t" the vector position is ⃗r1 and at "dt" after, it is ⃗r2; the resultant d⃗r as well as ⃗v (⃗v = d⃗r/dt) is tangent to the path along the vector unit ⃗eθ. d⃗r and ⃗eθ are parallel.

At the time "t", the velecity is ⃗v1 and at "dt" after, it is ⃗v2; the resultant d⃗v as well as ⃗a (⃗a = d⃗v/dt) is not tangent to the path, along a certain vector directed toward the center of curvature of the path. The acceleration vector can be decomposed to a component along the vector unit ⃗er (radial) and the second perpendicular vector unit ⃗eθ (tangential) .

Since the two vectors ⃗er and ⃗eθ are orthogonal , we can write:
⃗eθ . ⃗er = 0
The derivative gives:
d⃗eθ . ⃗er + ⃗eθ . d⃗er = 0
d⃗eθ = - ⃗eθ . d⃗er/⃗er
d⃗eθ = - eθ . der . ⃗er /er²
(The vectors ⃗eθ and ⃗er are collinear, thus: ⃗eθ . ⃗er = |⃗er| . |⃗eθ| = der . eθ)
We have der = er . dθ Then:
d⃗eθ = - eθ . er . dθ . ⃗er /er²
d⃗eθ = - (eθ/er) . dθ . ⃗er
d⃗eθ = - dθ . ⃗er
For unit vectors:
d⃗eθ = - (eθ/er) . dθ . ⃗er (4.1)

We have already set ⃗v = d⃗r/dt = (r dθ/dt)⃗eθ. Thus:
⃗a = d⃗v/dt = d[(r dθ/dt)⃗eθ ]/dt = (dr/dt) (dθ/dt)⃗eθ + (r d²θ/dt²)⃗eθ + (r dθ/dt) d⃗eθ/dt =
= [(dr/dt) (dθ/dt) + (r d²θ/dt²)]⃗eθ + (r dθ/dt) d⃗eθ/dt
The second term becomes, according to the relationship (4.1):
(r dθ/dt) d⃗eθ/dt = - (r dθ/dt) (dθ/dt) ⃗er
Hence:
⃗a = [(dr/dt) (dθ/dt) + (r d²θ/dt²)]⃗eθ - (r dθ/dt) (dθ ⃗er)/dt =
[(dr/dt) (dθ/dt) + (r d²θ/dt²)]⃗eθ - r (dθ/dt)² ⃗er

⃗a = [(dr/dt) (dθ/dt) + (r d²θ/dt²)] ⃗eθ - [r (dθ/dt)²] ⃗er
(component along ⃗eθ + component along the ⃗er)

6. General motion: particular cases:

6.1. Circular motion:

The motion is circular when the related tarjectory is a circle; that is the vector position ⃗r is constant in magnitude "r". The expression of the acceleration becomes:
⃗a = [(r d²θ/dt²)] ⃗eθ - [r (dθ/dt)²] ⃗er

ω = dθ/dt is called angular velocity of unit rad/s.

6.2. Uniform motion:

The motion is uniform when the velocity ⃗v is constant in magnitude "v"; that is the speed = ||⃗dr||/dt = r . dθ/dt = constant, which means dθ/dt = constant. We have then:
d²θ/dt²) = 0. Hence the expression of the acceleration becomes:
⃗a = [(dr/dt) (dθ/dt) - [r (dθ/dt)²] ⃗er

6.3. Circular Uniform motion:

The two above cases, combined give the expression of the acceleration for a circular uniform motion:
⃗a = - [r (dθ/dt)²] ⃗er

A circular uniform motion (CUM) has just a radial component: a = r (dθ/dt)².
Since : || ⃗v|| = v = r (dθ/dt), we have: a = r (dθ/dt)² = r (v/r)² = v²/r

We define the period of revolution T in seconds for a CUM as the period of time that the oject takes to complete a cycle: ω = Δθ/Δt = 2π/T . The related frequency ƒ in s^-1 or Hertz is the number of cycles comleted per second: T = ƒ .
We have also the relationships between the angular and tangential velocities ω and v v = r dθ/dt = r ω. "r" is the radius of the circle.
ω = 2π/T = 2πƒ
v = r ω

7. CUM Applications:

7.1. Gravity function of altitude:

The law of universal gravitation for the two objects (satellite and earth) of mass m_s and m_e (5.9742 × 10²⁴ kg) respectively is written as:
F = G m_s m_e / z²;
where G is the gravitational constant, approximately equal to 6.673×10^-11 N m2 kg^-2, and "z" is the distance betweem the two objects.

The distance "z", in this example is equal to the altitude of the satellite above Earth "h" plus the Earth radius r = 6,357 km : z = h + r

We assume that the motion of the satellite is circular and uniform. That yields:

F = G m_s m_e / z² = m_s g_h,
where g_h = g(z) is the acceleration of gravity at the distance z = r + h.
When h = 0, g(z) becomes g(0) = g₀ = 9.81 m s ^-2 (at Maryland). We have:
G m_s m_e / z² = m_s g_h --> G m_e / z² = g_h. Then:
g_h = G m_e / z² = G m_e / (r + h)²
g_h = G m_e / (r + h)²

G m_e = 6.673×10^-11 x 5.9742 × 10²⁴ = 4.0 x 10¹⁴
g₀ = g(0) = 4.0 x 10¹⁴ / (6,357.00 x 10³)² = 4.0 x 10¹⁴ / (6.357 x 10⁶)² = 4.0 x 10¹⁴ x 2.47 x 10^-12 = 9.90 m s^-2

7.2. Speed of a satellite:

If the satellite orbits at an altitude of 300 km, then:
g_h = g(300 km) = 4.0 x 10¹⁴ / (6,357.00 + 300.00)² = 4.0 x 10¹⁴ / (6.657 x 10⁶)² = 9.03 m s^-2 (less than g₀)

By equating the centripetal force m_s v²/z on the satellite and the weight m_s g_h of this object, we can write:
m_s a = m_s v²/z = m_s g_h. The inertial mass in the centripetal force, and the gravitational mass in the weight are equivalent, then:
v²/z = g_h v = [z x g_h] ^1/2 = [(r + h) x g_h] ^1/2

v (h) = [(r + h) g_h] ^1/2 = [G m_e / (r + h)] ^1/2

This tangential speed does not depend on the mass of the orbiting object. It is a function of the altitude h.

v = [(6,357.00 + 300.00) x 9.03 ] ^1/2 = [(6.657) x 9.03 x 10⁶] ^1/2 =
[(6.657) x 9.03] ^1/2 x 10³ = 7.75 x 10³ m.s^-2.

For the moon, the altitude h is about 385,000 km, then:
v (h) = [(r + h) g_h] ^1/2 = [4.0 x 10¹⁴ / (6,357 + 385,000)x 10³] ^1/2
= [4.0 x 10¹⁴ / (3.91) x 10⁸] ^1/2 = = [10⁶] ^1/2 = 10 ³ = 1 km/s.
v(moon) = 1 km/s
According to ω = 2π/T, and v = z ω the period of the moon is T = 2π/ω = 2π z / v = 2π x 391,370 km / 1 km/s = 2π x 391,370 s = 682.72 hours = 28.45 days.

7.3. Escape speed:

Un object thrown upward falls, because it was launched with a small speed, but with an escape speed, it will bot fall. How much does this speed be?
We assume that the origin of the potential energy is the earth surface. Then, the potential energy of object to launch is E_p = - m g₀ r
(m is the mass of the object, g_p the acceleration of the gravity at the surface of Earth, and "r" is the radius of Earth. At this position, the object is thrown at the velocity v_e, then its kinetic energy is E_k = (1/2) m v_e². The total energy of the object is
E_t = E_p + E_k = - m g₀ r + (1/2) m v_e²
E_t = - m g₀ r + (1/2) m v_e²

If the object escapes, it will approach a place where its total energy is null. As the total energy is conserved, E_t = 0. That is:
- m g₀ r + (1/2) m v_e² = 0. That yields:
g₀ r = (1/2) v_e². Then:
v_e = [2 g₀ r]^1/2

v_e = [2 g₀ r]² = [2 x 9.81 x 6,357 km]^1/2 = 11.2 km/s

We have then three cases for an object of speed v :
- v < v_e: The object falls.
- v = v_e: The object orbits.
- v > v_e: The object escapes.

7.4. Rounding corner:

A car rounds a corner od f radius r 50 m. The coefficient of static friction between the tires and the road is μ_s = 0.80. There is a limit for the speed of the car in the corner not to exceed to avoid skidding.

The three forces acting on the car: Gravity, the normal force N and the force of static friction. The force of static friction provides the centripetal force required for the car to keep it moving in the circular path. The greater the speed of the car, the more friction is needed.Hence:

mv²/r <= f_s = μ_sN =
μ_smg. That is:

v²/r <= μ_sg, or

v <= [μ_sgr]^1/2
The speed limit is given by (independent of mass):

v = [μ_sgr]^1/2

V = [0.80. 9.81.50]^1/2 = 20 m/s
Remark:
The sum of the components of force gives: Verically: Σ F_y = W - N = 0 or N = mg Horizontally: Σ F_x = f_s= μ_smg = mv²/r

7.5. Banked Curves:

Many roads are tilted or banked on corners by design. On a banked curve, the required centripetal force is given by the normal force N, that is the component N sin θ..
Without no need to friction, we have the two forces W and N acting on the car. The component N cosθ cancels the weight W:
Σ F_y = - Ncos θ + W = 0, so
N cos θ = mg
And Σ F_x = N sinθ = m a_x = mv²/r
Therefore, to remains in the circular motion:
mv²/r <= N sinθ = mg sinθ/N sinθ = mg tg θ That is :
v <= [g r tg θ]^1/2
The speed limit is :

v = [g r tg θ]^1/2

For a car, witout a need to the friction force rounding a corner in a circle of 80.0 m tilted at 45 ^o, the speed limit is:
v <= [9.81 . 80. 1 ]^1/2 = 28 m/s