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Mathematics Teaching Self-evaluation Tools

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For the following questions, select the statement which most accurately matches your level of confidence (1 is not confident and 4 is very confident) or choose from the alternatives detailed in the question. You do not have to answer all questions. Your answers will be saved so you can exit and come back to your self-evaluation at any time. Click Save and Results to view the next steps for questions you have answered.

1.

How confident are you that

1

2

3

4

a.

you understand the language of kinematics?

Define the following:

Position, displacement, distance; speed, velocity; acceleration, magnitude of acceleration; relative velocity (in 1 dimension).

Solution:

Position – describes the location of an object

Distance – describes how far something has moved

Displacement – describes how far something has moved, in a specific direction

Speed – describes a change in distance in a specific amount of time

Velocity - describes speed in a specific direction

Acceleration – describes a change in velocity

Magnitude of acceleration – describes the size of the acceleration (i.e. without a direction)

Relative velocity (in 1 dimension) – describing the velocity of one particle relative to another. For example – when driving at 30mph and another car overtakes, doing 10 mph faster – so its real speed is 40mph, but relative to your vehicle it is travelling at 10mph.

b.

you know the difference between position, displacement and distance?

On Saturday I left home and walked 4 km along a road heading due north. I then went 2km along a road heading due East.

a. Describe my final position
b. Describe the displacement of the second part of this journey.
c. What is the distance I have travelled in total?

Answers:

a. My final position is 4km due north, and 2km due East, from where I started
b. The displacement is 2km due East
c. I have travelled 6km.

These three concepts are very similar.

Position is a way of describing the location – using coordinates would be an example of position.

Distance is a way of measuring how far – so in this case just the 4km, the direction (north) is not needed.

Displacement is a vector quantity – it combines the distance with a direction – so 4km north.

c.

you know the difference between velocity and speed, and between acceleration and magnitude of acceleration?

Explain why a velocity of 4ms-1 due north is different from a speed of 4ms-1.

Find two answers to a velocity of 5ms-1 due north after an acceleration with magnitude 1ms-2 occurs.

Solutions:

A speed of 4ms-1 could be orientated in any direction – it could be north, but it could also be in any other direction A velocity of 4ms-1 due north must be in a northerly direction.

An acceleration with magnitude of 1ms-2 could mean the velocity has changed by +1 – so now being 6ms-1 due north, or -1 – so now being 4ms-1 due north.

It could also be changed by 1ms-1 in any other direction – so the velocity might now be 5ms-1 (north) plus 1ms-1 (east) – or i + 5j in vector form.

Advanced students might be able to understand thise locus below.

d.

you are able to draw and interpret kinematics graphs, knowing the significance (where appropriate) of their gradients and the areas underneath them?

Consider the velocity-time graph shown below.

a. What is the acceleration between O and A?
b. What is the distance travelled in the first 10 seconds?
c. What is the distance travelled between A & B.
d. What is the acceleration between B & C?
e. What is the total distance travelled?

Solutions:

a. The velocity changes from 0ms-1 to 20ms-1 in 10 seconds – so the acceleration is 2ms-2.

b. The distance travelled is calculated by finding the area under the graph. This part of the graph forms a triangle, so the area is found by calculating ½ x 10 x 20 = 100m.

c. The distance travelled is calculated by finding the area under the graph. This part of the graph forms a rectangle so the area is found by calculating 20 x 20 = 400m

d. The velocity changes from 20ms-2 to 0ms-1 in 5 seconds, so the acceleration is -4ms-2.

e. The total distance travelled needs the distance from B to C added to the two other distances already found.

B to C is a triangle with area = ½ x 5 x 20 = 50m.

So the total will be 100 + 400 + 50 = 550m.

Alternatively the whole shape can be calculated in one calculation – a trapezium – so ½ x 20 x (35 + 20) = 10 x 55 = 550m.

e.

you are able to differentiate position and velocity with respect to time and know what measures result?

The position of a particle is given by

a. Find the position when

b. Find the velocity when

c. Find the acceleration when

Solutions:

a. Substituting into the equation given shows

.

b. Finding will give an expression for the velocity.

This is sometimes written as .

So and when then .

c. Differentiating again, to give sometimes written as , gives the acceleration.

So and when then .

f.

you are able to integrate acceleration and velocity with respect to time and know what measures result?

A particle moves so that its acceleration is given by .

Given that the velocity is and the displacement is equal to zero when , find an expression for both the velocity and the displacement in terms of .

From we can integrate to give and then use the initial conditions to show that (since gives the velocity, which must be at ).

So

From we integrate again to give and then use the initial conditions to show that (since gives the displacement, which must be zero when ).

So .

g.

you are able to recognise when the use of constant acceleration formulae is appropriate?

A ball is held out of a window and then dropped. The ball falls freely under gravity and takes 5 seconds to reach the ground.

[take ]

How far did the ball fall?

How fast was the ball moving at the instant when it hit the ground?

We recognise that the acceleration of the ball is due to gravity, and will be constant. So we need the appropriate constant acceleration formulae.

Writing and then selecting

will give

Now using will give .

h.

you are able to solve kinematics problems using constant acceleration formulae and calculus?

Consider a stone dropped, from rest, from a height of 3m above a pond. The stone is dropped and falls into the water.

When it hits the water the acceleration is given by .

Calculate:

a. The velocity of the stone as it hits the water.

b. The velocity of the stone at a depth of 1m.

Solutions:

a. Is a constant acceleration problem – the stone will accelerate downwards thanks to gravitational attraction.

So writing and

using gives the velocity as .

b. Now in the water the situation changes to one needing calculus.

The acceleration is no longer constant and we can see that when , as here, the stone will decelerate.

Now we need to analyse the differential equation.

Starting to measure time from when the stone hits the water gives us

and when .

The differential equation can be written as

and so which gives .

Taking exponentials on both sides gives .

Since we have when then we find .

Thus at a depth of we have

which gives .

The graph of the solution is illuminating as to the behaviour of the stone in the water in this context: the rate of slowing down decreases as the speed approaches the terminal velocity –

so the speed is an exponential decay, towards . Kinematics Diagram 2 to be added here.