Linear Kinematics

PEP 294: Lecture Notes

X. Linear Kinematics

Kinematic Quantities (pp. 294-308)

1. Position

Position:

- the location of the object of interest at a given instance

- a vector quantity

- r = (x, y)

Examples: (4, 3), (-1, 7), (-10, -4)

Standard unit: m (meter)

2. Distance vs. Displacement

Distance:

- length of the path moved

- a scalar quantity: magnitude only

Displacement:

- change in position

- a vector quantity: magnitude + direction

- net effect of motion

d = r₂ - r₁ = (x₂, y₂) - (x₁, y₁) = (x₂ - x₁, y₂ - y₁)

d (magnitude of d) =

tan q = => q (direction of d) = tan^-1

Standard unit: m (meter)

Example: r₁ (initial position) = (4.0, 5.0) m, r₂ (final position) = (6.0, 2.0) m. Compute d (displacement).

d = (6.0, 2.0) - (4.0, 5.0) = (6.0 - 4.0, 2.0 - 5.0) = (2.0, -3.0) m

3. Velocity vs. Speed

Velocity:

- rate of change in position

- how fast an object moves in which direction?

- a vector quantity

- velocity =

v =

Speed:

- how fast an object moves

- a scalar quantity

- speed =

v =

Standard unit: m/s

Examples:

1. d = 0.9 km, = 0.5 hr => speed?

v = = 1.8 km/h

2. d = (2.0, -3.0) m, = 1.5 s => velocity?

v = (, ) = (1.33, -2.0) m/s

3. v_swim = 2 m/s, v_current = 0.5 m/s

v = 2.06 m/s, q = 14°

Cyclic movement:

- swimming or running

- velocity = cycle length (CL) * cycle rate (CR)

swimming: v = SL (m/stroke) x SR (strokes/s)

running: v = SL (m/stride) x SR (strides/s)

- Example: SL = 5.0 m/stride, SR = 2.0 strides/s

v = (5.0)(2.0) = 10.0 m/s

4. Acceleration

Acceleration:

- rate of change in velocity or speed

a = = (magnitude + direction)

a = = (magnitude only)

Standard unit: m/s²(= m/s/s)

Examples:

1. v₁ = 3.0 m/s, v₂ = 11.0 m/s, = 10 s => acceleration?

a = = = 0.8 m/s²

2. v₁ = (3.0, 4.0) m/s, v₂ = (4.0, 10.0) m/s, = 10.0 s

a = = = (0.1, 0.6) m/s²

3. a = -0.3 m/s², v₁ = 4 m/s, v₂ = 0 m/s => = ?

= = -4.0 / -0.3 = 13.3 s

Magnitude & direction:

- change in magnitude of velocity => acceleration

- change in direction of velocity => acceleration

a = 0 => no change in velocity (constant velocity)

Projectile Motion (pp. 309 - 327)

Projectile:

- object projected into the air

- Examples: balls, human body, arrow, etc.

- follows a parabolic path

1. Kinematic quantities of interest

- Horizontal displacement: shotput, discus, javelin, long jump, etc.
- Max. vertical displacement: high jump, pole vaulting, etc.
- Trajectory (accuracy & speed): basketball, soccer, etc.

2. Factors Affecting the Projectile Motion

Forces acting on a projectile:

- Gravity:

(1) pulls projectile downward

(2) causes parabolic trajectory

(3) gravitational acceleration: (0, -g), vertical acceleration only

where g = 9.81 m/s²

- Air Resistance:

(1) affects trajectory of the projectile

(2) normally ignored for simplicity

Initial conditions:

- projection velocity (v)

- angle of projection (q)

3. Properties of the Projectile Motion

Mass vs. Projectile Motion:

Free-Fall vs. Horizontal Projection:

Types of Projectile Motion:

4. Analysis of projectile motion

Strategies:

- resolve projection velocity into h & v components

- treat horizontal & vertical motions separately

- use equations of projectile motion:

Resolution of the projection velocity:

v_x = v·cos q

v_y = v·sin q

- Example (soccer ball): v = 15 m/s, q = 35° => initial H and V velocities?

v_x = v·cos q = (15)·(0.819) = 12.29 m/s

v_y = v·sin q = (15)·(0.574) = 8.60 m/s

Equations of Projectile Motion:

Vertical v_v = v_y - g·t
d_v = v_y·t -
v_v² = v_y² - 2g·d (where, g = 9.81 m/s²)

Horizontal v_h = v_x
d_h = v_x·t

Miscellaneous Ascending Time: T =
Height of the Apex: H =
Horizontal Distance: D = v_x·(2T)

- Example (soccer ball): v_x = 12.29 m/s, v_y = 8.60 m/s => positions & velocities?

After 0.1 s:

d_h = v_x·t = (12.29)(0.1) = 1.30 m

d_v = v_y·t - = (8.60)(0.1) - (0.5)(9.81)(0.1)²= 0.81 m

v_h = 12.29 m/s

v_v = v_y - g·t = 8.60 - (9.81)(0.1) = 7.62 m/s

After 0.2 s:

d_h = (12.29)(0.2) = 2.46 m

d_v = (8.60)(0.2) - (0.5)(9.81)(0.2)² = 1.52 m

v_h = 12.29 m/s

v_v = 8.60 - (9.81)(0.2) = 6.64 m/s

After 0.5 s:

d = (6.14, 3.07) m

v = (12.29, 3.70) m/s

After 1.0 s:

d = (12.29, 3.70) m

v = (12.29, -1.21) m/s

- Example (soccer ball): v_x = 12.29 m/s, v_y = 8.60 m/s => height of the apex, ascending time, & max. horizontal distance?

H = = = = 3.77 m

T = = = 0.88 s

D = = (12.29)(2)(0.88) = 21.63 m

Vertical	v_v = v_y - g·t d_v = v_y·t - v_v² = v_y² - 2g·d (where, g = 9.81 m/s²)
Horizontal	v_h = v_x d_h = v_x·t
Miscellaneous	Ascending Time: T = Height of the Apex: H = Horizontal Distance: D = v_x·(2T)