In a speed-time graph speed is always plotted on the vertical axis and time is always plotted on the horizontal. This represents the motion of a particle accelerating from a speed at time $0$, $u$, to a speed $v$ at time $t$.
The gradient of the line of the speed-time graph is the acceleration of the particle (for straight lines acceleration is constant). \begin{align} \text{Gradient of line} & = \frac{\text{change of velocity} }{\text{time} }, \\ & = \frac{v - u}{t}, \\ &= a. \end{align} The area under the speed-time graph is the distance the particle travels. \begin{align} \text{Shaded area} & = \left( \frac{u + v}{2} \right)t, \\ & = s. \end{align}
Distance-time graphs and Acceleration-time graphs can also be drawn for the motion of a particle, where time is always plotted on the horizontal axis.
A straight line on a distance-time graph represents that a particle has constant speed. Curvature suggests that a particle is accelerating or decelerating.
Worked Example: Speed-time graphs and acceleration-time graphs
A car drives along a straight road. The car accelerates uniformly from rest to a speed of $10 \mathrm{ms^{-1} }$ in $2T \mathrm{seconds}$. It then travels at a constant speed for $1 \mathrm{minute}$. The car then decelerates uniformly to rest in a further $8T \mathrm{seconds}$. Sketch a speed-time graph to illustrate the motion of the car. If the total distance travelled by the car is $800\mathrm{m}$, find the value of $T$ and sketch and acceleration-time graph to illustrate the motion of the car.
We sketch the following speed-time graph.
The area of a trapezium is \begin{align} s & = \frac{1}{2} (a+b) h, \\ & = \frac{1}{2} (60 + 2T + 60 + 8T) \times 10, \\ & = 5 (10T + 120) \mathrm{m}. \end{align} Where $a$ represents the shorter of the two parallel sides of the trapezium and $b$ represents the other. We are given that $s=800 \mathrm{m}$, so we equate this to the above expression and solve for $T$. \begin{align} 5 (10T + 120) & = 800, \\ 10T + 120 & = 160, \\ 10T & = 40, \\ T & = 4. \end{align} The acceleration in the first $2T=8 \mathrm{seconds}$ is given by \begin{align} a& = \frac{10}{8}, \\ & = 1.25 \mathrm{ms^{-2} }. \end{align} Here, for the first 8 seconds the $v$-coordinate increases by 10 as the $t$-coordinate increases by 8, this gives a positive answer. The acceleration in the last $8T=32$ seconds is given by \begin{align} a & = \frac{-10}{32}, \\ & = -0.3125 \mathrm{ms^{-2} }. \end{align} Here, in the last 32 seconds the $v$-coordinate decreases by 10 as the $t$-coordinate increases by 32. This gives a negative answer.
Test Yourself
Try our Numbas test on Linear measure - distance and length.
Try our Numbas test on Speed.
=Distance -Time Graphs= ''''Distance'''' is the total length travelled by an object. The standard unit is the ''''metre''''. A distance-time graph shows how far an object has travelled in a given time. Distance is plotted on the Y-axis (left) and Time is plotted on the X-axis (bottom). Below you can see that the object represented by the blue line has travelled 10m in 2s whereas the object represented by the red line has only travelled 4m in this time and is therefore travelling more slowly. [image://i.imgur.com/qKjSSIo.png] ''''Straight lines'''' on a distance-time graph tell us that the object is travelling at a '''constant speed'''. Note that you can think of a stationary object (not moving) as travelling at a constant speed of 0 m/s. On a distance-time graph, there are no line sloping downwards. A moving object is always ''''increasing'''' its total length moved with time. [image://i.imgur.com/m72vOWp.png?1] ''''Curved lines'''' on a distance time graph indicate that the speed is changing. The object is either getting faster = ''''accelerating'''' or slowing down = ''''decelerating''''. You can see that the distanced moved through each second is changing. ==Calculating Speed from a Distance-Time Graph== [image://i.imgur.com/SZupBpn.png?1] The average speed can be calculated for any part of a journey by taking the change in '''distance''' and dividing by the change in '''time''' for that part of the journey. You can even do this for a curved line where the speed is changing, just remember that your result is the '''average''' speed in this case. You may also notice that the formula for calculating speed is sometime written with small triangles '''Δ''' (the Greek letter delta) in front of '''d''' (distance) and '''t''' (time). The '''Δ''' is just short hand for '''"change in"'''. Therefore '''Δt''' means '''"change in time"''' [image://i.imgur.com/ULJd8Ds.png] ==Displacement== ''''Displacement'''' is the length between start and stop positions and includes a direction. Displacement is a vector quantity. [image://ocw.uci.edu/cat/media/OC08/11004/OC0811004_L6Graphic06.gif]If an object goes back to where is started in certain time, then its displacement is zero. Its distance would be the total length of the journey. A displacement-time graph is able to show if an object is going backwards or forwards. Usually, a line with a negative gradient would indicate motion going backwards. This cannot be shown on a distance-time graph. Image Source: //ocw.uci.edu/ ==Describing the motion of an object== In most mechanical problems we are asked to determine the connection between speed, position and time. Will two cars crash if they are heading towards each other as they apply brakes at a certain time? To describe the position of a moving object, you have to specify its position relative to a particular point or landmark that is understood by everyone. Along a straight line, you only need the position of the landmark and how far the object is from the landmark left or right (or east or west). 5 metres from the door does not mean anything without giving some indication of direction (inside or out for example). Now left or right is not a good distinction as not everyone can agree with it. In Physics, we specify the origin landmark at 0 and the points either side of it are either positive or negative numbers (in units of metres). [image://i.imgur.com/CGRPorR.png] When describing the motion of an object try to be as detailed as possible. For instance... During ''''Part A'''' of the journey the object travels '''+8m''' in '''4s'''. It is travelling at a '''constant velocity''' of '''+2ms^-1^''' During ''''Part B'''' of the journey the object travels '''0m''' in '''3s'''. It is '''stationary for 3 seconds''' During ''''Part C'''' of the journey the object travels '''-8m''' in '''3s'''. It is travelling at a ''''constant velocity'''' of ''''-2.7ms^-1^'''' back to its starting point, our reference point 0. Why can we use ''''velocity'''' instead of ''''speed''''? Because by labelling our two directions + and -, we now know which way our object is moving in 1-dimension, forwards or backwards.
The key point here is exactly the idea of the constant speed (rectangle).
Let it be an arbitrarily changing velocity. Since at every instant in time the speed changes, it will be almost impossible to calculate the displacement. At this point, the best you can do is to give away of the actual value and accept an approximated value. To do so, at certain moments in time, let’s say at every 2 seconds you pick the corresponding value of speed and calculate the displacement as if speed was constant. At this point, what you will end up with are small rectangles representing small amounts of the displacement. The total approximated displacement will be the sum of the small displacements.
To better show this, let’s use the triangle (constant acceleration) of your example. In the first figure the plot of displacement vs time for a particle moving with a constant acceleration of $\rm a=1m/s^2$ is shown. After $\rm t=10 s$ it forms the triangle with sides of 10 by 10 units, whose area is $\rm \frac{1}{2}×10×10$ represented in red.
After Appling the 2 by 2 seconds sampling process you get the green rectangles of the second image. Has you can see, the total approximated displacement is an underestimation of the actual displacement, since, for example, in $\rm t=3s$ you still consider that velocity equals $\rm 2m/s$ but in reality it is higher ($\rm 3m/s$).
However, if you make thinner and thinner rectangles and then sum them up, your approximation will get closer and closer to the actual value. For example, if one uses a 1 second sampling, the error pointed before for for $\rm t=3s$ will disappear, but not the error for $\rm t=2.5s$ or $\rm t=3.5s$.
In the end, if you use instant sampling (infinitely thin rectangles) and then sum the resulting infinite number of rectangles your approximated value will match the actual value and the green area (which we shown to be equal to the approximated displacement) will become equal to the red area, i.e, the actual value for the displacement equals the area under the speed vs time curve.
For further reading search for the following keywords.
Geometric representation of integration.
Integration as summation.
PS: This doesn’t mean displacement is a sort of “area”. It is just the result of representing graphs or plots of functions in a “piece of paper”. This means that when one represents any quantity as a graph the magnitude of such a quantity will be represented as a length, therefore, any quantity that results from the multiplication of the axes of such a graph will be represented as an area. For example, work will be equal to the area under the force vs displacement curve, and electric charge will be equal to the area under the current intensity vs time curve.