Kinematics: Motion of Particles and Bodies
Constant acceleration, projectiles, rotation, relative motion and velocity of mechanisms.
Position, velocity and acceleration
Kinematics describes motion without asking what causes it. Velocity is the rate of change of position, v = dx/dt, and acceleration is the rate of change of velocity, a = dv/dt = v dv/dx. The area under an acceleration–time graph is the change in velocity, and the area under a velocity–time graph is the distance traveled; the slope of a position–time graph is velocity.
Constant acceleration
v = v₀ + a t x = x₀ + v₀ t + ½ a t² v² = v₀² + 2 a (x − x₀)
Worked example: accelerating and braking
Worked example: a car from 0 to 100 km/h in 8 s
- 100 km/h = 100 ÷ 3.6 = 27.78 m/s. Acceleration a = 27.78 ÷ 8 = 3.47 m/s² (0.35 g).
- Distance in the 8 s: ½ a t² = ½ × 3.47 × 64 = 111 m.
- Braking from 100 km/h at a steady 8 m/s² (0.82 g, good dry tires): stopping distance v² ÷ 2a = 48.2 m in 3.47 s. Double the speed and the distance quadruples: 200 km/h needs 193 m.
- Dropping a part from a bench 1.5 m high: t = √(2h ÷ g) = 0.553 s and it lands at v = g t = 5.42 m/s (19.5 km/h).
Projectile motion
With no drag the horizontal speed is constant and the vertical motion is free fall: x = v cos θ · t and y = v sin θ · t − ½ g t². Range R = v² sin 2θ ÷ g, peak height H = v² sin² θ ÷ 2g and flight time T = 2 v sin θ ÷ g (launch and landing at the same height).
Worked example: a ball thrown at 20 m/s
- At 45°: vₓ = v_y = 14.14 m/s; flight time 2.88 s; range 40.8 m; peak height 10.19 m reached after 1.44 s.
- At 30° and 60° the range is the same, 35.3 m (check: 35.3 m), but the 60° throw peaks at 15.3 m against 5.1 m and takes 3.53 s against 2.04 s.
- Air drag shortens real trajectories and moves the best angle below 45°; for anything fast or light (a golf ball, a dart) numerical integration is needed (see Programming Languages).
Rotation about a fixed axis
ω = ω₀ + α t θ = ω₀ t + ½ α t² v = ω r at = α r an = ω² r
Worked example: a spindle coasting to rest
- A spindle at 1 500 rpm (ω = 157.08 rad/s) coasts to rest in 20 s with constant deceleration: α = -7.85 rad/s² and it turns 250 revolutions while stopping.
- A point 0.1 m from the axis moves at ω r = 15.7 m/s and feels a normal acceleration ω² r = 2 467 m/s² (252 g): this is why a small unbalance at speed makes a large force (see Flywheels, Balancing and Rotors).
Relative motion and mechanisms
Velocity of a point B relative to a point A on the same rigid link is vB/A = ω × rB/A, perpendicular to the line AB with magnitude ω times the distance. Adding relative velocities (vB = vA + vB/A) gives the velocity of every point of a linkage, as used for the slider-crank and four-bar analysis in Linkages and Mechanisms. The instantaneous center of rotation, the point with zero velocity at that moment, gives the same result geometrically.
Related
- Previous: Centroids, Area Moments and Friction. Next: Dynamics: Force, Energy and Momentum.
- Motion Control and Actuators for motion profiles.
- Engineering Fundamentals: the whole guide.
FAQ
Why does stopping distance grow with the square of speed?
Because v² = 2 a d: for a fixed deceleration the distance is proportional to v².
What is the difference between speed and velocity?
Speed is the magnitude; velocity is a vector with direction. Average speed over a closed path is not zero but the average velocity is.
How do I convert rpm to rad/s?
Multiply by 2π ÷ 60, about 0.1047. 1 500 rpm is 157.1 rad/s.
References
- Hibbeler, R. C., Engineering Mechanics: Dynamics, Pearson.
- Meriam, J. L., Kraige, L. G., Engineering Mechanics: Dynamics, Wiley.
- Norton, R. L., Design of Machinery, McGraw-Hill.
Last reviewed: October 2026. Examples ignore air drag unless stated.