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Scalar

A quantity with magnitude only, no direction

Vector

A quantity with magnitude and direction

Examples of Scalars

Mass, temperature, time, energy

Examples of Vectors

Displacement, velocity, acceleration, force

Resultant Vector (Perpendicular 3–4

Magnitude 5 units

Vector Components (Magnitude 10 at 30°

x=8.66, y=5.00

Distance

Total path length traveled, directionless

Displacement

Straight-line change in position from start to end with direction

Path vs. Displacement Example

3 m east then 4 m west → distance 7 m, displacement 1 m east

2D Displacement Example

3 m east then 4 m north → magnitude 5 m at 53.1° N of E

Position–Time Graph Slope

Velocity

Zero Slope on x–t Graph

Object at rest

Velocity–Time Graph Slope

Acceleration

Area Under v–t Graph

Displacement

Area Under a–t Graph

Change in velocity

Crossing v=0 on v–t Graph

Change in direction

Speed

Rate of motion without direction

Velocity

Rate of motion with direction

Average Speed Formula

Total distance ÷ total time

Average Velocity Formula

Displacement ÷ total time

Instantaneous Velocity

dx/dt

Instantaneous Acceleration

dv/dt

Acceleration

Rate of change of velocity over time

SI Unit of Acceleration

m/s²

Sign of Acceleration

Positive speeds up in + direction, negative speeds up in – direction

Deceleration

Acceleration opposite the velocity vector

Kinematic Eqn (No Δx

v=v₀+at

Kinematic Eqn (No v

Δx=v₀t+½at²

Kinematic Eqn (No t

v²=v₀²+2aΔx

Solve for a from v=v₀+at

a=(v−v₀

/t

Time from Δx=v₀t+½at²

t=[−v₀±√(v₀²+2aΔx

]/a

Unit Conversion 72 km/h→m/s

20 m/s

Relative Speed (Opposite Directions 20 m/s and 15 m/s

35 m/s

Free-Fall Distance in 3 s (g=9.8 m/s²

44.1 m

Free-Fall Final Speed after 3 s

29.4 m/s

Compute Average Speed

150 m in 20 s → 7.5 m/s

Piecewise Displacement

3 m/s for 4 s then −1 m/s for 6 s → 6 m

Instantaneous Velocity from x(t

=4t²−2t+1 at t=3 s

22 m/s

Final Speed (v₀=2 m/s, a=3 m/s², t=5 s

17 m/s

Displacement (v₀=2 m/s, a=3 m/s², t=5 s

47.5 m

Stopping Time (v₀=12 m/s, a=−3 m/s²

4.0 s

Stopping Distance (v₀=12 m/s, a=−3 m/s²

24 m

Speed from v²=v₀²+2aΔx (v₀=0, a=9.8, Δx=20 m

19.8 m/s

Average Acceleration (v

−5→15 m/s in 4 s

5.0 m/s²

Triangle Area on v–t (0→12 m/s in 6 s

36 m displacement

Instantaneous Accel from v(t

=6t²−2t at t=1 s

10 m/s²

Solve Time from Δx=v₀t+½at² (v₀=8, a=2, Δx=30

2.78 s

Projectile Time Up (v₀y=20 m/s, g=9.8

2.04 s

Projectile Total Time (same

4.08 s

Range at 45° (v₀=30 m/s, g=9.8

91.84 m

Round Trip Example (100 m out and back in 40 s

Average speed 5.0 m/s, average velocity 0 m/s

Piecewise Trip

2 m/s for 5 s, rest 3 s, −1 m/s for 4 s → displacement 6 m, distance 14 m

Instantaneous v from x(t

=10+5t−2t² at t=2 s

−3 m/s

Vector Add (Boat 3 m/s N, Current 4 m/s E

Ground speed 5.0 m/s at 36.9° N of E

Trapezoid Area on v–t (10→0 m/s over 5 s

25 m displacement

Convert 90 km/h→m/s

25 m/s

Speed from v²=v₀²+2aΔx (v₀=10, a=2, Δx=50

17.32 m/s