# Speed, velocity and acceleration

## In a nutshell

Speed is the distance travelled by an object per second of time. It is a scalar quantity and can be calculated using the distance-time equation. Acceleration is the change in velocity per second of time and is a vector quantity. Uniform acceleration is a constant acceleration.

**Equations**

#### Word equation | #### Symbol equation |

$speed = \dfrac{distance}{time}$ | $v = \dfrac{s}{t}$ |

$acceleration = \dfrac{change \ in \ velocity}{time}$ | $a = \dfrac{\Delta v}{t}$ |

$(final \ velocity)^2 - (initial \ velocity)^2 = 2 \times acceleration \times distance \ travelled$ | $v^2 - u^2 = 2 \times a \times s$ |

**Variable Definitions**

#### Quantity name | #### Symbol | #### Unit name | #### Unit |

$speed / velocity$ | $v$ | $metre\ per \ second$ | $m/s$ |

$distance$ | $s$ | $metre$ | $m$ |

$time$ | $t$ | $second$ | $s$ |

$acceleration$ | $a$ | $metre \ per \ second \ squared$ | $m/s^2$ |

$final \ velocity$ | $v$ | $metre \ per \ second$ | $m/s$ |

$initial \ velocity$ | $u$ | $metre \ per \ second$ | $m/s$ |

## Speed and velocity

Speed measures how fast an object is travelling. It is a scalar quantity. However, giving speed a direction makes it into a vector quantity.

##### Example

*A car travels at a speed of $20\,m/s$*. *The car travels at a velocity of $20\,m/s$ to the west.*

The speed an object travels at is defined as the distance an object travels per second of time. This gives the equation for speed as:

$speed = \dfrac{distance}{time}\\ \ \\ \\ \ \\v = \dfrac{s}{t}$

##### Example

*A bowling ball rolls down a bowling lane in $8\,s$ at a speed of $2.4\,m/s$ and knocks over all the pins! What is the distance of the bowling lane?*

*First, write down the values given in the question:*

*$t = 8\,s\\ \ \\ v = 2.4\,m/s$*

*Next, write down the equation needed and rearrange:*

*$v = \dfrac{s}{t} \\ \ \\ s = v \times t$*

*Then substitute the values into the equation:*

*$s = 2.4 \times 8 = 19.2\,m$*

*Make sure to include the correct units and round to $2\ s.f.$:*

*$s = 19\,m$*

*Therefore, the distance of the bowling lane is $\underline{19\,m}$.*

## Acceleration

If an object is not travelling at a constant speed, it must have an acceleration. An object with a positive acceleration will be speeding up and an object with a negative acceleration (i.e. a deceleration) will be slowing down. Acceleration is a vector quantity.

**Note: **There is acceleration when there is a change in velocity, which means there is a change in speed or direction!

The average acceleration can be found using the equation:

$acceleration = \dfrac{change \ in \ velocity}{time}\\ \ \\ a = \dfrac{\Delta v}{t}$

##### Example

*A car travelling at $15\,m/s$ comes to a stop at a traffic light. It takes $5\,s$ for the car to reach rest. Calculate the acceleration of the car.*

*First, write down the quantities given in the question:*

*$\Delta v = v_2 - v_1 = 0 - 15 = -15\,m/s$*

*Next, write down the equation needed:*

*$a = \dfrac{\Delta v}{t}$*

*Then substitute the values into the equation:*

*$a = \dfrac{-15}{5} = -3 \,m/s^2$*

*Make sure to include the correct units and round to $2\ s.f.$:*

*$\text{acceleration, }a = -3.0\,m/s^2$*

*The answer is given as a negative value as the car is slowing down (decelerating).*

*Therefore, the acceleration of the car is $\underline{-3.0\,m/s^2}$. *

### Uniform acceleration

A constant acceleration is referred to as uniform acceleration. To calculate uniform acceleration, use the equation:

$(final \ velocity)^2 - (initial \ velocity)^2 = 2 \times acceleration \times distance \ travelled \\ \ \\v^2 -u^2 = 2\times a \times s$

##### Example

*Acceleration due to gravity is roughly $10\,m/s^2$. If an object initially at rest is dropped from a helicopter, what will its velocity be after falling $0.2 \,km$? Give the velocity as a vector with a direction. Assume there are no resistive forces acting on the object.*

*First, write down the quantities given in the question:*

*$u = 0\,m/s\\ \ \\ a = 10\,m/s^2\\ \ \\ s = 0.2\,km = 200\,m$*

*Next, write down the equation needed and rearrange:*

*$v^2 - u^2 = 2\times a \times s\\ \ \\v^2 = u^2 + (2 \times a \times s)\\ \ \\v = \sqrt{u^2 + (2\times a \times s)}$*

*Then substitute the values into the equation:*

*$v = \sqrt{0^2 + (2\times 10 \times 200)} = 63.24... \,m/s$*

*Make sure to include the correct units and round to $2\ s.f.$:*

*$\text{velocity, }v = 63\, m/s$*

*Therefore, the final velocity is $\underline{63\,m/s}$ downwards.*