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

Variable Definitions

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.