Learning Lab

Displacement \(\left(x\right)\)

When an object is moved from one point to another it is said to be displaced. If we move to Dandenong we can say we have been displaced \(32\,km\) from Melbourne, but to define our position, we need to specify a direction and say we are \(32\,km\) South East of Melbourne. That is we give a magnitude (\(32\,km\)) and a direction (South East). Consequently displacement is a vector quantity.

Note that displacement is often referred to as position.

Distance \(\left(d\right)\)

Distance is the magnitude (size) of the displacement, but has no direction so it is a scalar quantity. For example you can say my distance is \(60\,km\) away, but your displacement is \(60\,km\) in a direction - such as \(60\,km\) West.

Velocity \(\left(v\right)\)

Velocity \(v\) is the rate of change of displacement \(x\). That is, for an object moving at constant velocity, \[\begin{align*} v & =\frac{x}{t} \end{align*}\] where \(t\) is the time taken. Since displacement is a vector quantity, so is velocity. The units of velocity depend on the units of displacement and time. If displacement is in metres \(\left(m\right)\) and time is in seconds \(\left(s\right)\) the unit of velocity is metres per second and is written as \(m/s\) or \(ms^{-1}.\) 11 Note that \(m/s\) is metres divided by seconds. The notation \(ms^{-1}\) uses indicial notation. Here \(s^{-1}=1/s\) and so \[\begin{align*} ms^{-1} & =m\times s^{-1}\\ & =m\times\frac{1}{s}\\ & =\frac{m}{s}. \end{align*}\]

Speed

The speed of an object is the magnitude (size) of it’s velocity. It has no direction and so is a scalar quantity.

Acceleration \(\left(a\right)\)

Acceleration is the rate of change of velocity. That is \[\begin{align*} a & =\frac{v}{t}. \end{align*}\] If \(v\) is in metres per second, the unit of acceleration is metres per second per second22 The notation \(ms^{-2}\) uses indicial notation. Here \(s^{-2}=1/s^{2}\) and so \[\begin{align*} ms^{-2} & =m\times s^{-2}\\ & =m\times\frac{1}{s^{2}}\\ & =\frac{m}{s^{2}}\\ & =m/s^{2}. \end{align*}\] and is written as \(m/s^{2}\) or \(ms^{-2}.\) Since velocity is a vector quantity, so too is acceleration.

Area Under Velocity Time Graph

Consider the units of the area. The area is the product of the velocity and time and so the units are \(m/s\,\times s=m.\) That is, the area under the graph gives displacement.

The area under a velocity-time graph gives displacement.

The area under a speed-time graph gives distance.

Consider the velocity time graph above. The area has been divided into three regions, \(1,2\) and \(3.\) Let area of each region be \(A_{1},A_{2}\) and \(A_{3},\)respectively. We have:44 Remember that the area of a triangle \(A_{t}\) is half \(\times\)base \(\times\) height and the area of a rectangle \(A_{r}\) is base \(\times\) height. That is, \[\begin{align*} A_{t} & =\frac{1}{2}\times b\times h\\ & =\frac{1}{2}bh\\ A_{r} & =b\times h\\ & =bh. \end{align*}\]

\[\begin{align*} A_{1} & =\frac{1}{2}\times b\times h\\ & =\frac{1}{2}\times20\times60\\ & =600\,m\\ A_{2} & =b\times h\\ & =10\times60\\ & =600\,m\\ A_{3} & =\frac{1}{2}\times b\times h\\ & =\frac{1}{2}\times30\times60\\ & =900\,m. \end{align*}\] The total displacement is therefore \(A_{1}+A_{2}+A_{3}=600+600+900=2100\,m.\)

Gradient of a Velocity-Time Graph

The gradient (or slope)55 The gradient \(m,\) of a linear (straight line) graph is the rise divided by the run. That is, \[\begin{align*} m & =\frac{\text{rise}}{\text{run}}. \end{align*}\] of a Velocity-Time graph gives the acceleration of the object.

Consider the velocity-time graph above. From \(A\) to \(B\) the gradient of the graph is \(m_{AB}\) where \[\begin{align*} m_{AB} & =\frac{\text{rise}}{\text{run}}\\ & =\frac{\text{change in velocity}}{\text{change in time}}\\ & =\frac{60-0}{20-0}\\ & =\frac{60}{20}\\ & =3\,ms^{-2}. \end{align*}\] That is, the car is increasing its speed by \(3\,m/s\) each second.

From \(B\) to \(C\) the gradient \[\begin{align*} m_{BC} & =\frac{\text{rise}}{\text{run}}\\ & =\frac{\text{change in velocity}}{\text{change in time}}\\ & =\frac{60-60}{30-20}=\\ & =\frac{0}{10}\\ & =0\,ms^{-2}. \end{align*}\] That is the car is not accelerating but is moving at a constant velocity of \(60\,m/s.\)

From \(C\) to \(D\) the gradient \[\begin{align*} m_{CD} & =\frac{\text{rise}}{\text{run}}\\ & =\frac{\text{change in velocity}}{\text{change in time}}\\ & =\frac{0-60}{60-30}=\\ & =-\frac{60}{30}\\ & =-2\,ms^{-2}. \end{align*}\] Note the negative value of acceleration indicates the car is slowing down. We say the car is decelerating.

Further examples

1. A car accelerates from a stationary position. The rate of acceleration is constant66 If acceleration is constant we also use the expression “the acceleration is uniform”. . Sketch graphs of the acceleration against time, the velocity against time and the displacement against time.

Solution:
The acceleration-time graph is shown below.

It is simply a horizontal line as the acceleration is constant.
The velocity-time graph is shown below.

It is a linear graph. The gradient (slope) of the graph is the acceleration.
The displacement-time graph is shown below.

This is a curve. The gradient (slope) at any point on the curve is the velocity at that time. You can find the approximate value of the gradient by drawing a tangent at the point of interest and then calculating the gradient of the tangent.77 If you know the functional form of the displacement, then you can differentiate it to find the gradient at any time. For example, if \[\begin{align*} x\left(t\right) & =3t^{2}-2t+1 \end{align*}\] then the gradient is \[\begin{align*} \frac{dx}{dt} & =6t-2. \end{align*}\]

2. A ball is thrown vertically upwards with a speed of \(5\,m/s\). Sketch and describe the graph of velocity of the ball against time from the moment it is thrown to the moment it touches the ground. Ignore air resistance.

Solution:
First define the coordinate system you want to work in. We take upwards velocity as positive and downwards velocity as negative. Once the ball is thrown up, the only force acting on it is gravity which is constant. So from the previous example, the graph of velocity against time is linear. That is, a straight line. As gravity acts downwards, the velocity against time graph must have a negative gradient (because the ball is decelerating). So a velocity-time graph looks like the one below.

Now we need to describe it. Think about what has happened. Initially the ball is thrown into the air vertically upwards (positive direction) with a speed of \(5\,m/s\). Point \(A\) describes the initial velocity (upward) that the ball has. It is \(5\,m/s\) upward.
Then the ball starts to slow down (due to gravity acting on it). This is the part of the graph from \(A\) to \(B\). At point \(B,\) the ball has zero velocity. It is not moving, and so this is the point of maximum altitude for the ball.
After \(B,\) the ball has a negative velocity. That means it is now heading downwards towards the point from which it was thrown. Point \(C\) represents the starting position. At this point the ball has the same speed88 Speed is the magnitude of the velocity vector. At point \(A\) the velocity was \(5\,m/s\) upward. At point \(C\) the velocity is \(-5\,m/s\) or \(5\,m/s\) downward. Remember, a vector quantity has to have a magnitude and a direction. as it had at point \(A.\) This is shown in the diagram below.

Summary

The relationships between various graphs and their gradient and the area below them are summarized in the table below: