e=mc2andallthat

Necessity may well be the mother of invention, but teacher desperation is often the mother of new pedagogy.

For an unconscionably long time, I think that I failed to adequately help students understand the so called ‘equations of motion’ (the mathematical descriptions of uniformly accelerated motion using the standard v, u, a, s and t notation) because I suffered from the ‘curse of knowledge’: I did not find the topic hard, so I naturally assumed that students wouldn’t either. This, sadly, proved not to be the case, even when the equation was printed on the equation sheet!

What follows is a summary of a dual coding technique that I have found really helpful in helping students become confident with problems involving the ‘equations of motion’. This is especially true at GCSE, where students encounter formulas such as

for the first time.

A dual coding convention for representing motion

Applying dual coding to an equation of motion problem

EXAMPLE: A car is travelling at 6.0 m/s. As the car passes a lamp-post it accelerates up to a velocity of 14.2 m/s over a distance of 250 m. Calculate a) the acceleration; and b) the time taken for this change.

The problem can be represented using the dual coding convention as shown below.

Note that the arrow for v is longer than the arrow for u since the car has a positive acceleration; that is to say, the car in this example is speeding up. Also, the convention has a different style of arrow for acceleration, emphasising that it is an entirely different type of quantity from velocity.

We can now answer part (a) using the FIFA calculation system.

Part (b) can be answered as shown below.

Visualising Stopping Distance Questions

These can be challenging for many students, as we often seem to grabbing numbers and manipulating them without rhyme or reason. Dual coding helps make our thinking explicit,

EXAMPLE: A driver has a reaction time of 0.7 seconds. The maximum deceleration of the car is 2.6 metres per second squared. Calculate the overall stopping distance when the car is travelling at 13 m/s.

We need to calculate both the thinking distance and the braking distance to solve this problem

The acceleration of the car is zero during the driver’s reaction time, since the brakes have not been applied yet.

We visualise the first part of the problem like this.

Now let’s look at the second part of the question.

There are three things to note:

• Since the car comes to a complete stop, the final velocity v is zero.
• The acceleration is negative (shown as a backward pointing arrow) since we are talking about a deceleration: in other words, the velocity gradually decreases in size from the initial velocity value of u as the car traverses the distance s.
• Since the second part of the question does not involve any consideration of time we have omitted the ‘clock’ symbols for the second part of the journey.

We can now apply FIFA:

Since the acceleration arrow points in the opposite direction to the positive arrow, we enter it as a negative value. When we get to the third line of the Fine Tune stage, we see that a negative 169 divided by negative 4.4 gives positive 38.408 — in other words, the dual coding convention does the hard work of assigning positive and negative values!

And finally, we can see that the overall stopping distance is 9.1 + 38.4 = 47.5 metres,

Conclusion

I have found this form of dual coding extremely useful in helping students understand the easily-missed subtleties of motion problems, and hope other science teachers will give it a try.

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If you have enjoyed this post, you might enjoy these also:

• Dual Coding SUVAT problems — mainly focused on KS5 content, but applicable for KS3 and KS4
• SHM and the Top Gear Challenge — mainly focused on the KS5 simple harmonic motion content