Displacement and velocity

In a nutshell

Displacement is the change of position of an object, and is a vector quantity so has both a magnitude and direction. Velocity is the rate of change of displacement. It is also a vector quantity as it is the speed in a given direction.

Equations

Variables

Speed

Average speed is the rate of change of distance. It can be calculated using the distance travelled and time taken:

​$$v=\dfrac{\Delta x}{\Delta t}$$​​

Instantaneous speed is the speed of an object over a very short time interval. It can be calculated by determining the gradient of a distance-time graph at a specific interval.

Velocity

Vector quantities have both magnitude and direction. Velocity is the rate of change of displacement. The displacement of an object is its distance in a specific direction, and is therefore a vector quantity. 

The velocity of an object is the rate of change of displacement. It is also a vector quantity, and can be expressed mathematically as:

​$$v = \dfrac{\Delta s}{\Delta t}$$

A velocity-time graph shows the velocity over a period of time. A slope shows acceleration, whereas a straight line shows constant velocity.

Example

London is about $$277\,km$$​ from Exeter, but is $$314\,km$$ by road. It take about 3 hours and 55 minutes to travel from London to Exeter by road. Calculate the average speed, then the average velocity during the journey. 

Calculating speed:

First, state the variables from the question:

​$$\Delta x = 314\,km \newline t = 55 \space minutes$$

Then, state the equation needed:

$$average\, speed, v = \dfrac{\Delta x}{\Delta t}$$

Next, convert time to seconds:

$$t = 3 \ hours \ 55 \ mins\newline \\[0.1in] t = (3 \times 60 + 55) \times 60\newline \\[0.1in] t = 3480 \ s$$​​

Then. substitute in the values and solve:

​$$average\, speed, v = \dfrac{314,000}{3480} \\[0.1in] average\, speed, v = 22\,ms^{-1}$$

Calculating velocity:

First, state the variables from the question:

​$$\Delta s = 277 \,km$$

Then, state the equation needed:​

​$$average\, velocity, v = \dfrac{\Delta s}{\Delta t}$$​​

Next, substitute in and solve:

​$$average\, velocity, v = \dfrac{277,000}{3480}\\[0.1in]average\, velocity, v = 20 \,ms^{-1}$$​​

​

The average speed from London to is $$\underline{22 \, ms^{-1}}$$ and the average velocity $$\underline{20 \, ms^{-1}}$$.

Displacement-time graphs

The motion of an object can be expressed graphically on a displacement time graph. On this graph, a flat line represents a stationary object, as its gradient and therefore velocity is zero, as shown in graph a. A line with a constant gradient represents an object with constant velocity as shown in graph b, and a curved line represents acceleration as shown in graph c.

The gradient of a displacement-time graph tells the velocity of an object. If the graph is curved, a tangent can be drawn to calculate the velocity at a specific time. 

Example:

The motion of a bicycle is shown on the displacement-time graph below. Calculate its velocity between $$10 \ s$$ and $$20 \ s$$. Give your answer to two significant figures.

First, state the variables in the question:

​$$\Delta s = (190- 35)\,km = 155 \,km\newline t=10\,s$$

Then, state the equation needed:

$$v = \dfrac{\Delta s}{\Delta t}$$

Then finally sub in and solve:

​$$v = \dfrac{155}{10}\newline \\[0.1in] v = 15.5 \ ms^{-1}$$​​

The bike is moving at a velocity of $$\underline{16 \ ms^{-1}}$$ to two significant figures.