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3.4.1.3 Motion along a straight line

Displacement, speed, velocity, acceleration.

$$v=\frac{\Delta s}{\Delta t}$$ $$a=\frac{\Delta v}{\Delta t}$$

Calculations may include average and instantaneous speeds and velocities.

Representation by graphical methods of uniform and nonuniform acceleration.

Significance of areas of velocity–time and acceleration–time graphs and gradients of displacement–time and velocity–time graphs for uniform and non-uniform acceleration eg graphs for motion of bouncing ball.

Equations for uniform acceleration:

$$v=u+at$$ $$s=\frac{u+v}{2}t$$ $$s=ut+\frac{1}{2}at^{2}$$ $$v^{2}=u^{2}+2as$$

Acceleration due to gravity, g.

The suvat equations

It is important, right at the start, to understand the differences between the scalar and vector quantities you will use when investigating motion.

Speed is the rate of change of distance whereas velocity is the rate of change of displacement.

$$\large v=\frac{\Delta s}{\Delta t}$$

As distance is a scalar quantity speed itself is scalar. Displacement is a vector, therefore velocity is a vector. This means that velocity, unlike sped, can be broken into components and must be stated with a direction.

As both speed and displacement are often represented by the letter s we will only now refer to velocity (v) and displacement (s), if we ever need to refer to speed, we will use the letter c.

Acceleration is the rate of change of displacement,

$$\large v=\frac{\Delta s}{\Delta t}$$

In practice Δv, the change in velocity is calculated by:

$$\large \Delta v=\left ( v-u \right )$$

Where v is the final velocity and u is the initial velocity.

From these basic assumptions it is possible to derive four equations which can be used to describe the motion of objects under a constant acceleration. The four, so called, suvat equations that you need to be able to use are given below:

(eq. 1)

$$\large v=u+at$$

(eq. 2)

$$\large s=\frac{\left ( u+v \right )}{2}t$$

(eq. 3)

$$\large s=ut+\frac{1}{2}at^{2}$$

(eq. 4)

$$\large v^{2}=u^{2}+2as$$

Although you do not need to know how to derive these equations, if you are interested, you can find out where they come from here:

How to derive the four equations of motion

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Graphing motion

It is often useful to be able to represent the motion of an object on a graph, the two most common being, displacement-time and velocity-time graphs, although we do sometimes use acceleration-time graphs too. It is certainly worth thinking about how any particular object’s journey would look like on all three.

The simplest of the three is the displacement-time graph. Using the same basic principles that were discussed above we can see that an s-t graph can be used to find both the velocity and the the displacement of an object. The displacement is read directly from the y axis, and as the gradient is $\frac{Δ s}{Δ t}$ it equals the velocity of the object. If the line is sloping upward, then the velocity is positive, and if it slopes downwards the velocity is negative. The steeper the line, the greater the magnitude of the velocity. When the object accelerates, as the velocity is constantly changing the line would be curved. To find the velocity at any point a tangent to the line must be drawn. This is a skill that you may be expected to perform.

Displacemt-time graphs can either be drawn like the one shown, or (as is often the case for oscillating objects) shown with a positive and a negative quadrant.

displacement time graph
Figure 1: A displacement-time graph.

You should spend some time looking at this graph and describing the motion at each stage of it before the lesson.

Velocity time graphs are even more useful, but often require some careful thought when drawing them. On v-t graphs the gradient is $\frac{Δ v}{Δ t}$ which we know is equal to acceleration. The area underneath any section of the line is equal to the displacement of the object.

velocity time graph
Figure 2: A velocity-time graph.

There are some important points to note:

  • Whenever an object stops moving (even instantaneously) the line will cross the x-axis.
  • A constant velocity produces a horizontal line.
  • Any object experiencing a constant acceleration (e.g. an object in free-fall) will produce a straight line.
  • When finding the area under a line it is usually best to divide the area into several regular polygons and add their individual areas. This is easier than counting squares! It is of course possible (if you know the equation of the line) to integrate, but this is well beyond what is expected from the A-level course.

You should spend some time looking at the graph above and think what the acceleration at each stage is, and how that would look if we plotted acceleration against time. You will be expected to do this in class.

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