Kinematics of Particles

Kinematics describes motion without asking what causes it: where a particle is, how fast it moves and how quickly its speed or direction changes. Position, velocity and acceleration are linked by v = ds/dt and a = dv/dt, and every kinematics problem is a careful use of those two definitions.

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Position, velocity and acceleration

A particle is a body whose size does not matter to the question, such as a vehicle on a long track or a part in flight. Pick an origin and a positive direction first. Velocity is the rate of change of position and acceleration is the rate of change of velocity, so each is the slope of the one before it and each is the area under the one after it.

Eliminating time gives a third relation, a·ds = v·dv. Use it when acceleration is given as a function of position, or when a problem gives speeds and distances but no times.

Constant acceleration

When a is constant, integrating once and twice gives the three familiar equations: v = v₀ + a·t, s = s₀ + v₀·t + ½·a·t², and v² = v₀² + 2·a·(s − s₀). They only hold while a is constant. A motion with several phases, such as speed up, cruise and slow down, is solved one phase at a time, with the end of each phase as the start of the next.

Curvilinear motion

In two or three dimensions, velocity and acceleration are vectors. Rectangular components suit projectiles: horizontal motion has zero acceleration and vertical motion has a = −g, and the two share only the time. Normal and tangential (n–t) components suit motion along a known path: the tangential acceleration aₜ = dv/dt changes the speed, and the normal acceleration aₙ = v²/ρ, toward the centre of curvature, changes the direction.

Kinematics feeds the rest of dynamics. Once you know the acceleration, Newton’s second law gives the forces; the work–energy principle connects speed to position without finding the acceleration at all.

Key formulas

Formulas used on this page, with the textbook each is cited from
FormulaExpressionWhereCited from
Velocity and accelerationv = ds/dt, a = dv/dt, a·ds = v·dvs = position (m), v = velocity (m/s), a = acceleration (m/s²), t = time (s)Hibbeler, Engineering Mechanics: Dynamics
Constant accelerationv = v₀ + a·t, s = s₀ + v₀·t + ½·a·t², v² = v₀² + 2·a·(s − s₀)v₀, s₀ = values at t = 0; a constantHibbeler, Engineering Mechanics: Dynamics
Projectile motionx = x₀ + v₀ₓ·t, y = y₀ + v₀ᵧ·t − ½·g·t²g = 9.81 m/s², air resistance neglectedBeer et al., Vector Mechanics for Engineers: Dynamics
Normal and tangential componentsaₜ = dv/dt, aₙ = v²/ρ, a = √(aₜ² + aₙ²)ρ = radius of curvature of the path (m)Meriam & Kraige, Engineering Mechanics: Dynamics

Worked examples

Example: Travel time of an automated guided vehicle

Given

  • An automated guided vehicle (AGV) moves 30 m between two stations, starting and ending at rest.
  • It accelerates at 0.5 m/s² to a top speed of 1.5 m/s, cruises, then brakes at 0.75 m/s².

Find

  • The distance used by each phase.
  • The total travel time.

Solution

  1. Acceleration phase, from v = v₀ + a·t and v² = v₀² + 2·a·s.t₁ = 1.5 / 0.5 = 3.0 s, s₁ = 1.5² / (2 × 0.5) = 2.25 m
  2. Braking phase, with the same equations and a = −0.75 m/s².t₃ = 1.5 / 0.75 = 2.0 s, s₃ = 1.5² / (2 × 0.75) = 1.50 m
  3. Cruise phase covers the rest of the distance at constant speed. It is positive, so the AGV does reach top speed.s₂ = 30 − 2.25 − 1.50 = 26.25 m, t₂ = 26.25 / 1.5 = 17.5 s
  4. Add the three phases.t = 3.0 + 17.5 + 2.0 = 22.5 s

Answer: The AGV uses 2.25 m to speed up, 26.25 m cruising and 1.50 m to stop, for a travel time of 22.5 s. If the cruise distance had come out negative, the move would be a triangle profile that never reaches 1.5 m/s.

Example: A part leaving the end of a conveyor

Given

  • Parts leave the end of a horizontal belt conveyor at 1.2 m/s.
  • The belt surface is 0.9 m above the floor of a collection bin. Air resistance is neglected.

Find

  • How long a part is in the air.
  • How far past the belt end it lands.
  • Its speed and angle at impact.

Solution

  1. Vertical motion starts with v₀ᵧ = 0 and a = −g. Solve 0.9 = ½·g·t² for the flight time.t = √(2 × 0.9 / 9.81) = 0.428 s
  2. Horizontal motion has no acceleration.x = 1.2 × 0.428 = 0.514 m
  3. Vertical velocity at impact.vᵧ = 9.81 × 0.428 = 4.20 m/s
  4. Combine the components.v = √(1.2² + 4.20²) = 4.37 m/s at tan⁻¹(4.20 / 1.2) = 74° below horizontal

Answer: Each part is airborne for about 0.43 s, lands about 0.51 m past the belt end, and hits at about 4.4 m/s, 74° below horizontal. The landing point is set by belt speed and drop height, so the bin must move if either changes.

Common mistakes

In industry

Frequently asked questions

What is the difference between kinematics and kinetics?

Kinematics describes motion: position, velocity and acceleration. Kinetics relates that motion to the forces that cause it, through Newton’s second law, work and energy, or impulse and momentum.

When can I treat a body as a particle?

When its rotation and size do not affect the answer, so every point on it moves the same way. A car on a long road or a part in free flight can usually be a particle; a spinning wheel cannot.

Which constant-acceleration equation should I use?

Pick the one that contains your three known quantities and the unknown. If time is neither given nor asked for, v² = v₀² + 2·a·Δs avoids it.

When are n–t components better than x–y components?

When the path is known, such as a curved track or a rotating arm. Normal and tangential components separate the change in speed from the change in direction.

Keep learning

All Dynamics lessons · All engineering subjects

See also

Sources

  1. R. C. Hibbeler, Engineering Mechanics: Dynamics, Pearson
  2. F. P. Beer, E. R. Johnston, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Dynamics, McGraw Hill
  3. J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Dynamics, Wiley

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