Acceleration

Definition

Acceleration

Acceleration is the rate of change of velocity.

Definition

Deceleration

Deceleration is acceleration that reduces the magnitude of an object's velocity.

Acceleration is a vector quantity because velocity is a vector. A change in speed, direction or both produces acceleration.
The acceleration equation is
$a=v−uta=\frac{v-u}{t}$
where $a$ is acceleration in $m/s2\text{m/s}^2$ , $u$ is initial velocity in $m/s\text{m/s}$ , $v$ is final velocity in $m/s\text{m/s}$ and $t$ is time in $s\text{s}$ .
The unit $m/s2\text{m/s}^2$ means that velocity changes by that many metres per second during each second.
With one direction chosen as positive, a negative acceleration points in the negative direction. It represents slowing down only when velocity points in the positive direction.

Example

A car increases its velocity from $m/s6.0\ \text{m/s}$ to $m/s22\ \text{m/s}$ in $s4.0\ \text{s}$ .
Substitution gives $a=22−6.04.0a=\dfrac{22-6.0}{4.0}$ .
The acceleration is $m/s24.0\ \text{m/s}^2$ .

Rearranging the acceleration equation

Useful rearrangements are
$v = u + a t$
$u = v - a t$
$t=v−uat=\frac{v-u}{a}$
Keep the signs of velocities and acceleration during substitution. A negative value may carry essential directional information.

Example

A train travelling at $m/s25\ \text{m/s}$ slows uniformly at $m/s2-0.50\ \text{m/s}^2$ for $s30\ \text{s}$ .
Using $v = u + a t$ gives $v = 25 + (- 0.50) (30)$ .
The final velocity is $m/s10\ \text{m/s}$ .

Uniform acceleration without time

Definition

Uniform acceleration

Uniform acceleration is acceleration that remains constant.

When acceleration is uniform and time is not given, use
$v^2-u^2=2ax$
where $x$ is displacement in $m\text{m}$ . This equation applies only when acceleration is constant.
Useful rearrangements include
$x=v2−u22ax=\frac{v^2-u^2}{2a}$
$a=v2−u22xa=\frac{v^2-u^2}{2x}$
When solving for velocity, take the square root only after evaluating $v^2$ . Select the root that matches the stated direction of motion.

Example

A vehicle moving at $m/s20\ \text{m/s}$ comes to rest with uniform acceleration over $m50\ \text{m}$ .
Substitution gives $0^2-20^2=2a(50)$ .
Rearrangement gives $m/s2a=\dfrac{-400}{100}=-4.0\ \text{m/s}^2$ .

Interpreting acceleration

Positive and negative signs indicate direction relative to the chosen positive direction. They do not automatically mean speeding up and slowing down.
An object speeds up when its velocity and acceleration have the same sign.
An object slows down when its velocity and acceleration have opposite signs.
An object moving at constant speed in a circle accelerates because its direction, and therefore its velocity, continually changes.

Common Mistake

Do not use $m/s\text{m/s}$ as the unit of acceleration. The correct unit is $m/s2\text{m/s}^2$ .
Do not use $v^2-u^2=2ax$ when acceleration changes during the motion.
Do not discard a negative acceleration. Explain its direction or identify it as deceleration only when the motion direction makes that description correct.

Exam technique

Write the selected equation before substituting and keep all quantities in SI units.
Edexcel calculation marks commonly reward substitution, evaluation and the final unit, so show these stages clearly even when your calculator gives the answer directly.
Check whether the question gives distance or displacement and whether the signs make physical sense.

Self review

What does an acceleration of $m/s23\ \text{m/s}^2$ mean?
What equation links $u$ , $v$ , $a$ and $t$ ?
When may $v^2-u^2=2ax$ be used?
When does a negative acceleration make an object speed up?

Acceleration

Definition

Acceleration

Acceleration is the rate of change of velocity.

Definition

Deceleration

Deceleration is acceleration that reduces the magnitude of an object's velocity.

Acceleration is a vector quantity because velocity is a vector. A change in speed, direction or both produces acceleration.
The acceleration equation is
$a=v−uta=\frac{v-u}{t}$
where $a$ is acceleration in $m/s2\text{m/s}^2$ , $u$ is initial velocity in $m/s\text{m/s}$ , $v$ is final velocity in $m/s\text{m/s}$ and $t$ is time in $s\text{s}$ .
The unit $m/s2\text{m/s}^2$ means that velocity changes by that many metres per second during each second.
With one direction chosen as positive, a negative acceleration points in the negative direction. It represents slowing down only when velocity points in the positive direction.

Example

A car increases its velocity from $m/s6.0\ \text{m/s}$ to $m/s22\ \text{m/s}$ in $s4.0\ \text{s}$ .
Substitution gives $a=22−6.04.0a=\dfrac{22-6.0}{4.0}$ .
The acceleration is $m/s24.0\ \text{m/s}^2$ .

Rearranging the acceleration equation

Useful rearrangements are
$v = u + a t$
$u = v - a t$
$t=v−uat=\frac{v-u}{a}$
Keep the signs of velocities and acceleration during substitution. A negative value may carry essential directional information.

Example

A train travelling at $m/s25\ \text{m/s}$ slows uniformly at $m/s2-0.50\ \text{m/s}^2$ for $s30\ \text{s}$ .
Using $v = u + a t$ gives $v = 25 + (- 0.50) (30)$ .
The final velocity is $m/s10\ \text{m/s}$ .

Uniform acceleration without time

Definition

Uniform acceleration

Uniform acceleration is acceleration that remains constant.

When acceleration is uniform and time is not given, use
$v^2-u^2=2ax$
where $x$ is displacement in $m\text{m}$ . This equation applies only when acceleration is constant.
Useful rearrangements include
$x=v2−u22ax=\frac{v^2-u^2}{2a}$
$a=v2−u22xa=\frac{v^2-u^2}{2x}$
When solving for velocity, take the square root only after evaluating $v^2$ . Select the root that matches the stated direction of motion.

Example

A vehicle moving at $m/s20\ \text{m/s}$ comes to rest with uniform acceleration over $m50\ \text{m}$ .
Substitution gives $0^2-20^2=2a(50)$ .
Rearrangement gives $m/s2a=\dfrac{-400}{100}=-4.0\ \text{m/s}^2$ .

Interpreting acceleration

Positive and negative signs indicate direction relative to the chosen positive direction. They do not automatically mean speeding up and slowing down.
An object speeds up when its velocity and acceleration have the same sign.
An object slows down when its velocity and acceleration have opposite signs.
An object moving at constant speed in a circle accelerates because its direction, and therefore its velocity, continually changes.

Common Mistake

Do not use $m/s\text{m/s}$ as the unit of acceleration. The correct unit is $m/s2\text{m/s}^2$ .
Do not use $v^2-u^2=2ax$ when acceleration changes during the motion.
Do not discard a negative acceleration. Explain its direction or identify it as deceleration only when the motion direction makes that description correct.

Exam technique

Write the selected equation before substituting and keep all quantities in SI units.
Edexcel calculation marks commonly reward substitution, evaluation and the final unit, so show these stages clearly even when your calculator gives the answer directly.
Check whether the question gives distance or displacement and whether the signs make physical sense.

Self review

What does an acceleration of $m/s23\ \text{m/s}^2$ mean?
What equation links $u$ , $v$ , $a$ and $t$ ?
When may $v^2-u^2=2ax$ be used?
When does a negative acceleration make an object speed up?

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Rearranging the acceleration equation

Uniform acceleration without time

Interpreting acceleration

Review 2.2.2 Acceleration and the uniform-acceleration equation by teaching Genie

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Rearranging the acceleration equation

Uniform acceleration without time

Interpreting acceleration

Review 2.2.2 Acceleration and the uniform-acceleration equation by teaching Genie

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Rearranging the acceleration equation

Uniform acceleration without time

Interpreting acceleration

Review 2.2.2 Acceleration and the uniform-acceleration equation by teaching Genie

2.2.2 Acceleration and the uniform-acceleration equation Revision Guide

2.2.2 Acceleration and the uniform-acceleration equation

Acceleration

Rearranging the acceleration equation

Uniform acceleration without time

Interpreting acceleration

Review 2.2.2 Acceleration and the uniform-acceleration equation by teaching Genie

2.2.2 Acceleration and the uniform-acceleration equation Revision Guide