3.02MechanicsCore

Displacement, Velocity and Acceleration

Q: When do I use calculus instead of suvat?

Whenever acceleration is not constant — for example when s, v or a is given as a function of t.

Mechanics begins by describing motion precisely. $\textbf{Displacement}$ is how far an object is from a starting point in a given direction, $\textbf{velocity}$ is the rate at which displacement changes, and $\textbf{acceleration}$ is the rate at which velocity changes. Getting the distinctions right — distance versus displacement, speed versus velocity — prevents most errors before any equation is used.

45 min Kinematics

The big picture

The three kinematic quantities are linked by rates of change, so they are linked by calculus. Differentiate displacement to get velocity; differentiate velocity to get acceleration; integrate to go back. On graphs, those same relationships appear as gradients and areas: the gradient of a displacement–time graph is velocity, the gradient of a velocity–time graph is acceleration, and the area under a velocity–time graph is displacement. When acceleration is constant, this collapses into the suvat equations of the next lesson; when it varies, calculus is the only tool. Mechanics also introduces modelling language — particles, SI units, and assumptions — that every later topic uses.

What you'll be able to do

Distinguish vector and scalar quantities: displacement and distance, velocity and speed
Use SI units and interpret motion graphs using gradients and areas
Use $v = \dfrac{\mathrm{d}s}{\mathrm{d}t}$ and $a = \dfrac{\mathrm{d}v}{\mathrm{d}t}$ for variable motion
Use integration to find velocity and displacement, with initial conditions

Quantities, units and modelling

A $\textbf{vector}$ has size and direction; a $\textbf{scalar}$ has size only. Displacement, velocity and acceleration are vectors. Distance and speed are scalars.

In one dimension, direction is shown by sign: choose a positive direction and stick to it. A velocity of $-4$ m s⁻¹ means 4 m s⁻¹ in the negative direction.

SI units: displacement in metres (m), time in seconds (s), velocity in m s⁻¹, acceleration in m s⁻². Convert km h⁻¹ by multiplying by $\dfrac{1000}{3600}$ .

Objects are usually modelled as $\textbf{particles}$ — their size is ignored, so all their mass acts at a point and rotation can be neglected.

Tip — Before starting any mechanics problem, draw a diagram and mark the positive direction with an arrow. Most sign errors disappear.

Motion graphs

On a $\textbf{displacement–time}$ graph, the gradient is the velocity. A horizontal line means the object is at rest.

On a $\textbf{velocity–time}$ graph, the gradient is the acceleration and the $\textbf{area}$ between the graph and the time axis is the displacement. Area below the axis counts as negative displacement.

To find total $\textit{distance}$ from a velocity–time graph, add the areas above and below the axis as positive quantities.

Example 1A car accelerates uniformly from rest to 20 m s⁻¹ in 8 s, travels at constant speed for 30 s, then decelerates uniformly to rest in 12 s. Find the total distance travelled and the deceleration.

1The velocity–time graph is a trapezium.

2Area

= \tfrac12 \times 8 \times 20 + 30 \times 20 + \tfrac12 \times 12 \times 20 = 80 + 600 + 120

3Deceleration: gradient of the final section

= \dfrac{-20}{12}

AnswerDistance

800

m; deceleration

\tfrac53 \approx 1.67

m s⁻².

Variable acceleration: differentiation

When displacement is given as a function of time, velocity and acceleration come from differentiation: $v = \dfrac{\mathrm{d}s}{\mathrm{d}t}$ and $a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}$ .

An object is $\textbf{instantaneously at rest}$ when $v = 0$ . Its displacement is a maximum or minimum at that moment.

The maximum velocity occurs where $a = 0$ .

v = \dfrac{\mathrm{d}s}{\mathrm{d}t}, \qquad a = \dfrac{\mathrm{d}v}{\mathrm{d}t}

Rates of change.

Example 1A particle moves in a straight line with

s = 2t^{3} - 15t^{2} + 24t

metres for

t \ge 0

. Find when it is at rest and its acceleration at those times.

v = 6t^{2} - 30t + 24 = 6(t - 1)(t - 4)

, so at rest when

t = 1

and

t = 4

a = 12t - 30

3At

t = 1

a = -18

. At

t = 4

a = 18

AnswerAt rest at

t = 1

s (

a = -18

m s⁻²) and

t = 4

s (

a = 18

m s⁻²).

Variable acceleration: integration

Going the other way, $v = \int a\,\mathrm{d}t$ and $s = \int v\,\mathrm{d}t$ . Each integration introduces a constant, found from initial conditions such as "starts from rest at the origin".

Displacement between two times is $\int_{t_{1}}^{t_{2}} v\,\mathrm{d}t$ . If the velocity changes sign in that interval, split the integral to find the total distance.

s = \int v\,\mathrm{d}t, \qquad v = \int a\,\mathrm{d}t

Reversing the rates.

Example 1For the particle above, find the total distance travelled in the first 4 seconds.

v > 0

for

0 < t < 1

and

v < 0

for

1 < t < 4

, so split at

t = 1

s(1) - s(0) = 2 - 15 + 24 = 11

s(4) - s(1) = (128 - 240 + 96) - 11 = -16 - 11 = -27

4Distance

= 11 + 27

Answer

38

m (while the displacement at

t = 4

is only

-16

m).

Why this works

Displacement after 4 seconds is $-16$ m, yet the particle has travelled 38 m: it went forward 11 m, turned, and came back 27 m. Whenever velocity changes sign, displacement and distance part company.

Think like an examiner

State a positive direction and use signs consistently.
Distinguish distance from displacement, and speed from velocity, in your final answers.
Split integrals where velocity changes sign when finding distance.
Include units: m, m s⁻¹, m s⁻².

Common misconceptions

Giving the displacement when the question asks for distance.

Distance adds the magnitudes of forward and backward motion.

Reading displacement from the gradient of a velocity–time graph.

Gradient gives acceleration; area gives displacement.

Forgetting the constant when integrating acceleration.

Use initial conditions to find each constant.

Using a negative speed.

Speed is the magnitude of velocity and is never negative.

Kinematics

v = \tfrac{\mathrm{d}s}{\mathrm{d}t}

Velocity.

a = \tfrac{\mathrm{d}v}{\mathrm{d}t} = \tfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}

Acceleration.

s = \int v\,\mathrm{d}t

Displacement from velocity.

Stretch yourself

A particle starts from rest at the origin and moves in a straight line with acceleration $a = 6 - 2t$ m s⁻² for $t \ge 0$ . Find its maximum velocity, the time it next comes to rest, and its displacement at that time.

Hint — Integrate twice using the initial conditions. Maximum velocity occurs when $a = 0$ .

Reveal solution

SolutionIntegrating,

v = 6t - t^{2} + c

; at rest at

t = 0

gives

c = 0

, so

v = 6t - t^{2}

. Maximum velocity when

a = 0

, i.e.

t = 3

v = 18 - 9 = 9

m s⁻¹. It is next at rest when

6t - t^{2} = 0

, so

t = 6

s. Integrating again,

s = 3t^{2} - \tfrac13t^{3} + d

, and

s(0) = 0

gives

d = 0

. At

t = 6

s = 108 - 72 = 36

m. Since

v \ge 0

throughout

0 \le t \le 6

, the distance travelled is also 36 m.

\blacksquare

Questions students ask

What is the difference between distance and displacement?

Displacement is position relative to the start, including direction. Distance is the total length of path travelled, always positive.

When do I use calculus instead of suvat?

Whenever acceleration is not constant — for example when

s

v

a

is given as a function of

t

What does a negative acceleration mean?

Acceleration in the negative direction. It slows an object moving positively but speeds up one moving negatively.

Why model objects as particles?

It simplifies the mathematics by ignoring size and rotation, which is a good approximation when the object is small relative to the distances involved.

Key takeaways

Displacement, velocity and acceleration are vectors; distance and speed are scalars.
Gradient of $s$ – $t$ is velocity; gradient of $v$ – $t$ is acceleration; area under $v$ – $t$ is displacement.
Differentiate $s$ to get $v$ , and $v$ to get $a$ .
Integrate to reverse, using initial conditions for constants.
Split at sign changes of $v$ to find distance.