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MUNSCI helps you learn at your own pace, with student-led resources. Start now with courses in Mathematics AA/AI, Physics, Chemistry, Biology and Computer Science.
We've made hours worth of material in questions for the occasional question in a queue or for the midnight study sessions. Study how you want to - no judgement.
Our pride and joy. The resources you will find in the lessons are detailed walkthroughs including all our questions and dictionary terms integrated for an all around helpful study experience - designed by students.
We've made hours worth of material in questions for the occasional question in a queue or for the midnight study sessions. Study how you want to - no judgement.
Our pride and joy. The resources you will find in the lessons are detailed walkthroughs including all our questions and dictionary terms integrated for an all around helpful study experience - designed by students.
With a passion for all things science, we did our best to wrap up our knowledge and experience in the IB and MYP to bring helpful and meaningful resources for study and understanding to IB students.
Every Christmas (\({24}^{th}\) December) since 2003, Tanaka invested €\(d\) in a savings account to purchase a present one day for her younger brother, also born on Christmas day, 2003.
She continued to deposit €\(d\) on the \({24}^{th}\) of every month from then.
With a fixed rate of 0.32% interest per month, the interest was calculated on the last day of each month and subsequently added onto the account.
Let €\(D_{n}\)be the total amount in Tanaka's account on the last day of the \({n}^{th}\) month immediately after the interest has been added.
(a) Find an expression for \(D_{1}\) and show that \(D_{2} = {1.0032}^{2}d + 1.0032d\).
(b) (i) Find a similar expression for \(D_{3}\) and \(D_{4}\).
(ii) Hence, show that the amount in Tanaka's account the day before her brother's \({7}^{th}\) birthday is given by \(313.5({1.0032}^{83} - 1)d\).
(c) Write down an expression for \(D_{n}\) in terms of \(d\) on the day before the \({9}^{th}\) Christmas.
(d) Tanaka desired for the amount in her account to be at least €\(17\ 000\) the day before her brother turned 9 years old. Determine the minimum value of the monthly deposit required to achieve this. Give your answer correct to the nearest euro.
(e) As soon as the \({9}^{th}\) Christmas passed, Tanaka decided to invest €\(7\ 329\) of this money in an account of the same type earning 0.44% interest per annum. She withdraws €\(470\) every year on her Christmas day to buy her brother a present. Determine how many years and months it will take until there is no money in the account.
1. Construct a truth table [6] for:
2. Construct a logic gate for (A AND B) XOR C. [3]
3. User training is important. a) Outline why a company might choose to undertake user training over handing out user documentation? [5]
b) Which type of user training is the cheapest? [2] c) What might make a company choose remote/online classes? [3]
...and Mathematics
Not necessarily everyone loves Mathematics, but we love teaching it. Ever had issues with IB Mathematics? We sure did. We created the resource that we and other students were longing for.
Solving the quadratic using GDC (by graphing to find roots or through polynomial solve function):
\[PR=24.578 \text{ or } PR=-17.570\]
Reject the negative value (\(PR=-17.570\)), as a geometric line in 2D space cannot have a negative length.
Therefore:
\[PR=\boxed{24.6 \text{ cm}}\]
METHOD 2
The sine rule permits us to find \(α\) in the following sketch (not to scale, and may not reflect the true nature of the triangle, but provides a good approximate geometrical representation):
Using sine rule, we can instead attempt to find another angle in the triangle; by finding \(α\), we can find \(θ\).
Therefore using GDC by changing from radians into degrees:
Having explored vectors and scalars in topic 1, it can be applied in topic 2.1. Displacement and distance are extremely similar, but they have a key difference.
Distance (generally denoted by the variable \(d\)) is a scalar, which measures how far apart two locations are, along a traveled path.
Displacement (denoted by the variable \(\vec{s}\)) is a vector, which measures the length and the direction of the straight line joining two locations.
{{displacement}}
Speed, velocity, and acceleration
Again, both speed and velocity are very similar, but they are different.
Speed is a scalar, which measures the rate of change of distance. Mathematically:
\[\text{Speed} = { \Delta d \over \Delta t }\]
where \(d\) denotes distance, and \(t\) denotes time.
Velocity (denoted by the variable \(\vec{v}\)) is a vector, which measures the direction, and rate of change of displacement. Mathematically:
\[v = {\Delta s \over \Delta t}\]
where \(s\) denotes displacement, and \(t\) denotes time.
Acceleration (denoted by the variable \(\vec{a}\)) is another vector, which is kind of like the next step up from velocity. It measures the direction, and rate of change of velocity itself. Mathematically:
\[a = {\Delta v \over \Delta t}\]
where \(s\) denotes displacement, and \(t\) denotes time.
Acceleration is actually quite an integral part of IB physics, notably in topic 2, where there is an acceleration value which you must memorize by heart. This is the acceleration due to gravity of any free-falling object (on earth). This value is given by:
\[a = g = 9.81 \text{m}{\text{s}}^{-2}\]
Keep in mind that this value does not depend on the mass of the object itself.
Acceleration is a vector quantity, thereby making it very important to consider the direction in which it is acting. For instance, if the upwards (↑) direction is assumed to be positive (+), with \(g = 9.81 \text{m}{\text{s}}^{-2}\), then the acceleration due to gravity in the downwards (↓) direction would have a negative (-) value: \(g = -9.81 \text{m}{\text{s}}^{-2}\). Trust me, this becomes clearer through doing a bunch of practice problems.
Graphs describing motion
There are a total of 3 kinds of topic 2 graphs you may encounter in your IB exam. These are the displacement-time, velocity-time, and acceleration-time graphs.
Given these graphs, finding either the gradient of the line(s) or the area under the line(s) will give important values regarding motion, which are asked in both Paper 1 and Paper 2 exams.
The line’s gradient is velocity, as \(v = {\Delta s \over \Delta t}\)
The area under the line does not provide any useful information
The line’s gradient is acceleration, as \(a = {\Delta v \over \Delta t}\)
A straight but sloped line implies constant acceleration or deceleration
The area under the line is the displacement, as \(v \times t = s\)
The line’s gradient does not provide any useful information (for the IB syllabus)
A horizontal line implies constant acceleration (or deceleration if under the \(x\)-axis)
The area under the line is the velocity, as \(a \times t = v\)
The SUVAT equations of motion
In the data booklet at 2.1, there are a total of 4 given equations, with the variables:
\[s \rightarrow \text{displacement}\]
\[u \rightarrow \text{initial velocity}\]
\[v \rightarrow \text{final velocity}\]
\[a \rightarrow \text{acceleration}\]
\[t \rightarrow \text{time}\]
hence the name SUVAT. These equations relate these variables, in order to find unknowns when dealing with motion in the real world (ignoring air resistance of course).
Given certain values in a question, you can use these equations to determine other variables. For example, given time and speed, the acceleration and displacement can be identified through these equations.
Before getting into how to use these equations, it must be understood how these equations were derived.
The first SUVAT equation, \(v = u + at\), is relatively simple to derive. Constructing a velocity-time graph illustrating an object moving at constant acceleration:
The gradient of the line provides the acceleration, as
\[\text{Gradient} = {\Delta v \over \Delta t} = a\]
Considering the fact that \(\Delta v = \text{Final velocity} - \text{Initial velocity}\):
Then, simply rearranging the equation to isolate \(v\):
\begin{align*}a & = {v - u \over t} \\v - u & = at \\& \rightarrow \boxed{v = u + at} \\\end{align*}
The next SUVAT equation, \(s = ut + {1 \over 2}a{t}^{2}\), can be derived from the same velocity-time graph, but looking at the area under the line instead of the gradient:
The area under the curve outputs the total displacement \(s\), of the object. Splitting the area into 2 parts: \(s_1\) and \(s_2\) allows us to find the total displacement easily without the use of calculus:
The third SUVAT equation, \(s = ut + {1 \over 2}a{t}^{2}\) can be found by combining the last two equations. Simply substituting the first equation into the second for \(v\), and simplifying, leads to the third equation:
Lastly, the final SUVAT equation, \(v^2 = u^2 + 2as\), can be found by combining the same two equations, but rearranging for \(t\) instead of \(v\); slightly more complicated:
\[v = u + at \rightarrow ①\]
\[s = {1 \over 2}(u + v)t \rightarrow ②\]
Rearranging ① to isolate \(t\) as the subject:
\begin{align*}v - u & = at \\t & = {v - u \over a} \rightarrow ③ \\\end{align*}
The best way to demonstrate how to use these equations is through some practice and worked solutions/examples:
{{ q-p-aa06 }}
Projectile Motion
The motion an object undergoes when one does a free-throw in basketball, or simply leaping from one place to another, are all examples of projectile motion; an object following a curved path due to the influence of gravity. A simple example of projectile motion would look something like this:
Do keep in mind that in the IB exams, air resistance is usually negligible in calculations. This is because it would complicate things heavily, and it is much easier to use the symmetry provided by neglecting air resistance, for approximations. Nevertheless, you can be asked to draw or interpret graphs or situations where air resistance istaken into account:
Looking at individual components, specifically vertical and horizontal displacement, distance, and velocity, we can also identify differences when air resistance is negligible compared to when it is present:
The main points of consideration in projectile motion when air resistance is taken into account are:
Maximum height \(h\) (vertical displacement) is lower
Range \(R\) (horizontal displacement) is shorter
Trajectory is not symmetric
On the other hand, when air resistance is negligible:
Horizontal component of velocity is constant
Vertical component of velocity accelerates downwards at \(a=g=9.81 \text{m}{\text{s}}^{-2}\) (on Earth).
Projectile reaches maximum height \(h\) (vertical displacement) when its vertical velocity \(v_V\) is 0.
Trajectory is symmetric
Additionally, it must also be kept in mind that air resistance limits the maximum velocity an object can obtain from free-fall, due to the collisions with air particles essentially slowing down descent (this phenomenon is important in topic 2.2: Forces).
To solve problems involving projectile motion, SUVAT equations can be used. However, given the differences in horizontal and vertical components in projectile motion, SUVAT equations can only be used when the situation is resolved into horizontal and vertical components as in Topic 1– treating the horizontal and vertical components in the same way will not lead to the correct solution!
The way in which velocity changes in both horizontal and vertical directions can be exemplified:
Clearly, the horizontal component remains the same (air resistance is negligible), but the vertical component is constantly changing (due to the acceleration caused by gravity). Therefore, the components must be resolved before doing any SUVAT calculations.
Keeping that in mind, some facts regarding the SUVAT equations in each component can be noteworthy:
Again, the best way to understand how to apply these concepts is to go through questions with worked solutions.
