Mathematics AI HL

Sample Internal Assessment

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Table of content

Introduction

Rationale

Background information

Parameters of motion

Velocity – time graph

Resistive force

Angle of swirl after diving into the pool

Process of calculation

Trajectory of diver outside the pool

Sample calculation for 5m

Trajectory of diver inside the pool

Horizontal motion

Vertical motion

Determination of time period of the diver

Determination of depth reached by the diver

Determination of trajectory of the swimmer while diving

Trajectory of swimmer before getting into the pool

Trajectory of swimmer after getting into the pool

Case study

Conclusion

Evaluation

Bibliography

Swimming and diving have always been a staple in my life. From competitive swimming to diving off of 30 feet diving boards and even scuba diving, it has always been a recreational activity which I enjoy. These are part of some of the most practiced activities throughout the world; however, many people don’t realize the mathematical implications of these activities nor do they ascertain the many forces and variables acting simultaneously during these activities.

Since childhood, I am an avid swimmer. This has always interested me towards the study of science and mathematics related to this domain. Though the availability of resources on the trajectory of shot put, javelin, bullets and other such projectiles have been so extensive over the internet these days yet the need of the hour in context to mathematical modelling of swimming is rare to found. Since, the construction of swimming pool involves locus of the diver and modelling of motion involved in diving into the pool for a safe dive, mathematics and physics play a great role here. Recently while working on a problem to find the minimum depth that a swimming pool must have, I realized that knowing the trajectory of the driver is important to do this.

Understanding the mathematics behind the trajectory of any ‘body’ involves an extensive use of vectors, coordinate geometry, algebraic equation solving and manymore. To simplify the matter, I have determined the motion of a diver if he dives straight into the pool and continues to move vertically downward. Using the trapezoidal rule of integration(employed due to its simplicity and ease), the displacement – time equation for the diver has been determined. From different videos in the internet, I found that, the motion of a diver inside a swimming pool is more of a curved path than vertically straight downward. For a proper understanding of the idea with derived equations, I have gone through a few tutorials of buoyancy, effect of angle in trajectory of motion and research journals on the trajectory of projectile in presence of resistance. Finally, I decided to to model the trajectory of motion of a swimmer while diving into a swimming pool and determining the minimum depth of the pool for a safe diving based on the diver’s terminal velocity as my exploration in Mathematics Internal Assessment. This will be achieved through the application of vector algebra, coordinate geometry and conventional equation solving.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

The required terminologies are explained in brief as follows:

**Displacement:** It is the shortest distance between two locations (or positions) in a particular direction.

In the above figure, the black line is the displacement between the two positions ‘A’, &‘B’. Though the blue and the magenta line also joins the two positions ‘A’ & ‘B’, they are not displacement as they do not indicate the shortest distance between A and B.

**Velocity:** It is the rate of change of displacement of any ‘body’ or the change in displacement in unit time. It is usually measured in *m∙s*^{-1}.

**Acceleration:** It is the rate of change of velocity of any ‘body’ or the change in velocity in unit time. It is usually measured in *m∙s*^{-2}.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

This section intends to explain how the velocity of the diver changes with time from the time he starts to dive and till the point his body touches the pool. Here, the diver has been assumed to be a point object and the figure below shows how the velocity of the diver changes with time from the time he jumps for diving and moves vertically downward to the time he touches the pool.

As the initially the diver is at rest, initial velocity of the diver is zero and thus the graph passes through the origin. The graph resembles a straight line equation, y = ax, where y is the velocity of the diver and x is time. The gradient a represents the acceleration due to gravity which is pulling the diver down to the pool. As the diver approaches the pool, the velocity of the diver increases linearly (increases uniformly) due to the acceleration due to gravity acting on the diver.

As the velocity would increase linearly, the generalized equation of straight line has been considered of the form:

*v(t) = at + u ... *(1)

In the above equation (1), the notations are defined as:

*v*(*t*) = *terminal velocity*

*a = slope of curve or acceleration due to gravity*

*t = time*

*u = Y - intercept or initial velocity*

In this case, the diver is at rest before he starts. So, the initial velocity of the diver is 0. Therefore, *u = *0. This is in support of the fact that as shown in Figure - 2, the graph passes through the origin.

Thus, *v*(*t*) = *at *... (1)

Integrating the above equation (1) with respect to time over a time period from 0 seconds to *t* seconds:

\(\displaystyle\int\limits^t_0v(t)dt=\displaystyle\int\limits^t_0(at)dt\)

\(s=a\bigg[\frac{t^2}{2}\bigg]^t_0\)

\(s=\frac{1}{2}at^2\) ...(2)

Thus, equation (2) is the displacement -time equation of an object in a particular direction provided that the body is moving vertically downwards.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

A resistive force will be encountered by the diver once he enters into the pool. The resistive force, in this case, is the buoyant force exerted by the water in the swimming pool on the diver in the upward direction which provides an upthrust to the swimmer. The expression of buoyant force is mentioned below:

*F _{b} = Vρg...*(3)

In the above equation (3), the notations are defined as:

*F _{b} = Resistive buoyant force*

*V = Volume of water displaced while getting into the pool*

*ρ = Density of water*

*g = acceleration due to gravity*

As*, mass *(*m*)* = density *(*ρ*) *× Volume *(*V*)

According to Archimedes principle the volume of water displaced is equal to the volume of the object entering into the pool. Thus, here in equation-3, V can be considered as the volume of the diver.

*F _{b} = Vρg = mass *(

This means that the weight of the diver is acting on the diver in the upward direction and thus opposes the motion of the diver.

Angle of swirl is the angle at which the diver’s feet are kept while diving into the pool, which is not less than 15° with respect to the vertical axis (XX’). With a decrease in the angle, curvature of the trajectory of the diver inside the pool would be less which might result in causing injury due to striking with the base of the pool. On the contrary, an increase in angle may produce a huge thrust (resistive force on the diver) on the feet of the diver causing the diver to break his feet.

The trajectory of the diver could be divided into two parts because the motion outside water and that inside is fundamentally distinct due to variations in the forces acting on the diver.

They are –

- Outside the pool
- Inside the pool.

The motion in this part is vertically downward from the diving level to the pool. Due to my long association with swimming, I have observed that the diving platforms are generally at a height of 5m, 7.5m or 10m.

The equation of motion could be expressed as (from equation 2: s = \(\frac{1}{2}at^2\)):

\(s=\frac{1}{2}at^2\)

The body goes downward due to the acceleration due to gravity in the downward direction. Therefore, a = acceleration due to gravity = 9.81 m s^{-2}

Therefore, the displacement equation would be:

\(s(t)=\frac12×9.81×t^2\)

\(s(t)=4.9t^2 ...(4)\)

By differentiating equation (4) with respect to time ‘*t*’:

\(\frac{ds}{dt}=\frac{d}{dt}\) (4.9t^{2})

*v* (*t*) = 2 × 4.9 ×* t*

*v* (*t*) = 9.8*t*…(5)

This differentiation yields the function of velocity in terms of time showing the variation of velocity with respect to time.

Using equation (4), and (5), the terminal velocity of the diver while he touches the water surface would be calculated for three different diving platforms:

To do this, a primary data was collected by myself. I have visited three different pools with heights – 5.0 m, 7.5 m and 10.0 m. I dived into the pool and recorded the time taken for me to reach the surface of the pool using a water proof stop-watch. For each pool, three different trials were recorded.

Height (m)

Time 1 (sec)

Time 2 (sec)

Time 3 (sec)

Average Time (sec)

Velocity (m s^{-1})

5.00

1.02

0.97

1.01

1.00

9.8

7.50

1.13

1.26

1.21

1.20

12.1

10.00

1.42

1.37

1.41

1.40

14.0

From equation (4), the time taken by the diver to reach the water surface which is 5 m below the diving platform can be determined:

s = 4.9*t*^{2}

5 = 4.9*t*^{2}

*t*^{2 }= 1.02

t ≈ ±1

Since, time cannot be negative, hence the value of the time ‘*t*’ would be:

*t = *1* second*

From equation (5), the velocity of the diver at time *t = *1 second would be determined:

*v = *9.8*t*

*v = *9.8*×*1 = 9.8

It has been obtained from this section, that the velocity of the diver while getting into the pool would be maximum when the diver dives from the diving platform of 10m. Despite the obvious fact that more the height, more would be the terminal velocity more, there are certain flaws in the calculation that should be addressed. Some of the divers jumps from the diving platform which increases the diving height and hence the terminal velocity (velocity at the point when the diver touches the surface of the pool). Furthermore, in case of any wind current, the trajectory as well as terminal velocity would be vary.

The aim of this exploration is to determine the minimum depth of the swimming pool to ensure no injury of the swimmer due to collision with the floor of pool while diving into the pool. It is obvious that more the velocity while getting into the pool, the diver would go deeper into the pool. Hence, in this exploration, the velocity of the diver while diving from a height of 10 m would be considered to estimate the minimum depth of the swimming pool.

In this section of the trajectory, once the driver has entered the water surface, two forces are acting on the swimmer. They are: i) Driving force or the force due to gravity (body weight) and ii) Resistive force or the buoyant force.

From figure 5, it could be understood that, as the acceleration due to gravity acts in vertically downward direction and the buoyant force acts in vertically upward direction, the net force acting on the body is the difference between the two forces. As, the diver is considered to be floating, the buoyant force acting upwards is greater than the weight of the body acting downward.

*F _{net }= F_{b }- mg*

*F _{net }= Vρ_{w}g - Vρ_{b}g*

(since *ρ = *\(\frac{m}{V}\) where *m *is mass and *ρ* is density and *V *is volume)

*F _{net }= V_{g}(*ρ

From Newton’s 2^{nd} Law of Motion, the force applied on anybody is equal to the product of mass and acceleration of the body.

Using the principle, equation (6) could be reframed as:

*ma _{net }= Vg(*ρ

*V*ρ* _{b}a_{net }= Vg(*ρ

ρ* _{b}a_{net }= *ρ

*a _{net }= g *\(\bigg(\frac{ρ_w}{ρ_b}-1\bigg)\)...(7)

*a _{net} = net acceleration of the body inside the water*

*m = mass of swimmer in air*

*g = acceleration due to gravity*

*F _{b} = buoyant force*

*ρ _{w} = density of water*

*ρ _{w} = density of human body swimmer*

*V = volume of human body swimmer*

From figure 5, it should be noted that the direction of net force obtained is upward direction. Hence, the direction of anet is also upward. Throughout the exploration so far, the downward direction has been considered as positive and consequently, the upward direction would be considered as negative. Therefore,

*a _{net} = - g*\(\bigg(\frac{ρ_w}{ρ_b}-1\bigg)\)

*a _{net} = - g*\((1-\frac{ρ_w}{ρ_b})\)...(8)

As the diver goes into the pool, due to the angle made with respect to the vertical axis by his feet, the trajectory changes from vertically downward to a curl. As a result, in this section of the trajectory, the diver executes a two-dimensional motion – a vertical motion along a direction making an angle of with the vertical axes and a horizontal motion.

The motion in both the dimensions occur simultaneously. This is because when the diver tends to move along a given direction (say *x *axis), he entails a motion along the *y *axis as well. Hence, both will be considered separately.

As the body enters into the swimming pool, it has an initial velocity directed at a certain angle with the vertical axis. This has resulting in the following resolution of velocity vectors (resolution meaning splitting of vectors into components along the two axes yielding the given vector under observation).

sin *sin *θ = \(\frac{AB}{OB}\)

sin *sin *θ =\(\frac{v_x}{v}\)

*v _{x} *= sin

cos *cos** *θ = \(\frac{OA}{OB}\)

cos *cos** *θ = \(\frac{v_y}{v}\)

*v _{y} *=

As *v _{y} *represents the velocity along the

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

From the equation (2), the displacement time equation for the horizontal motion would be:

\(s_x=\frac{1}{2}a_xt^2+v_xt\)

From equation (8), it is clear that there is no acceleration component acting along the horizontal direction. Hence, the acceleration along the horizontal direction is zero.

*s _{x} = *\(\frac{1}{2}\)(0)

*s _{x }= v sin sin *θ

Where,

*s _{x }= range of Displacement along horizontal axis*

*v _{x} = Velocity along horizontal axis*

*a _{x} = Acceleration along horizontal axis*

As *s _{x} *represents the range of displacement along the horizontal axis, it also hence indicates, linking back to the exploration, how much water is needed for safe diving as the amount of water displaced needs to be less than the volume of the pool to a certain degree to prevent hazards.

From the equation (2), the displacement time equation for the vertical motion would be:

s_{y} = \(\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)\)*t ^{2} + (v cos cos θ) t…*(12)

Using equation (11) and equation (12), the maximum depth reached by the diver would be determined. First the body will go downward to reach the bottom of the pool which constitutes the first half of the trajectory covered by the body (diver) and during the second half of the trajectory, the body will move upward to reach the surface of water. As the acceleration in the upward direction is constant due to the buoyant force, time spent in the first half of the trajectory and in the second half remains equal. Hence, it could be assumed that, if the swimmer enters into the swimming pool and comes back to the surface in time *T*, then the swimmer would be at the maximum depth position at time \(\frac{T}2\). Hence, the time taken by the swimmer to reach the surface of the water after entering into the swimming pool should be determined.

From equation (12), it could easily be understood that at the surface of the water, the value of *s _{y} = *0

\(0=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρ_b}\bigg)t^2+(v\ cos\ cos\ θ)t\)

\(t\bigg[\frac{1}{2}\bigg(1-\frac{ρ_w}{ρ_b}\bigg)t+(v\ cos\ cos\ θ)\bigg]=0\)

Therefore:

\(\frac{1}{2}\bigg(1-\frac{ρ_w}{ρ_b}\bigg)\) *t* + (*v* cos *cos *θ) = 0

\(\frac{1}{2}\bigg(1-\frac{ρ_w}{ρ_b}\bigg)\) *t* = - *v* cos *cos *θ

\(\bigg(1-\frac{ρ_w}{ρ_b}\bigg)\) *t* = - 2*v* cos *cos *θ

\(\bigg(1-\frac{ρ_w}{ρ_b}\bigg)\) *t* = 2*v* cos *cos *θ

\(t=\frac{2vcoscosθ}{\bigg(\frac{ρ_w}{ρ_b-1}\bigg)}\)

Here, two values of time have been obtained because, the diver is at the surface of the water twice. Once, while he dives into the swimming pool and the other time when he gets out of the swimming pool. The first condition is the initial case where the time *t = 0.* Hence, the time period of the diver is:

\(t=\frac{2vcoscosθ}{\bigg(\frac{ρ_w}{ρ_b-1}\bigg)}\)

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

As mentioned earlier, during the first half of the trajectory, the diver reaches the bottom of the pool and the time taken for this is half of the total time (T) taken by the diver to enter the pool and come back to the surface. Thus, the maximum depth reached by the diver would be the vertical displacement of the diver at time \(t=\frac{T}{2}.\) Therefore, using equation 12, the maximum depth reached by the diver would be:.

\(s_y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)\) t^{2} + (*v* cos *cos *θ)*t*

\(s_y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)\bigg[\frac{2vcoscos\ θ}{2\big(\frac{ρ_w}{ρb}-1\big)}\bigg]^2\) + (*v* cos *cos *θ) \(\bigg[\frac{2vcoscos\ θ}{2\big(\frac{ρ_w}{ρb}-1\big)}\bigg]\)

\(s_y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)\bigg[\frac{vcoscos\ θ}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]^2\) + (*v* cos *cos *θ) \(\bigg[\frac{vcoscos\ θ}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]\)

\(s_y=-\frac{1}{2}\bigg(\frac{ρ_w}{ρb}-1\bigg)\bigg[\frac{vcoscos\ θ}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]^2\) + (*v* cos *cos *θ) \(\bigg[\frac{vcoscos\ θ}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]\)

\(s_y=\frac{1}{2}\bigg[\frac{(vcoscos\ θ)^2}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]^2\) + \(\bigg[\frac{(vcoscos\ θ)^2}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]\)

\(s_y=\frac{1}{2}\bigg[\frac{v^2θ}{\big(\frac{ρ_w}{ρb}-1\big)}\bigg]\) ...(13)

The trajectory of the swimmer could be expressed in two different parts. They are – i) Before getting into the pool, and ii) After getting into the pool.

From equation 4:

y = 4.9*t*^{2}

Where,

*y = displacement along vertical axis*

*t = time*

Therefore, from the equation, it could be said that during this part of the motion, there is no horizontal displacement.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

From the above study and from equation (11) and (12), it is clear that:

*s _{x} = v sin sin θ × t*

\(t=\frac{s_x}{vsinsin\ θ}\)

\(s_y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)t^2+(v\ cos\ cos\ θ)t\)

\(s_y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)\bigg(\frac{s_x}{vsinsin\ θ}\bigg)^2\) + (*v* cos *cos *θ) \(\bigg(\frac{s_x}{vsinsin\ θ}\bigg)\)

\(y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρb}\bigg)\bigg(\frac{x}{vsinsin\ θ}\bigg)^2\) + *x* cot cot θ …(14)

Where,

*y = displacement along vertical axis*

*x = displacement along horizontal axis*

Therefore, the trajectory of the swimmer is parabolic in nature. It should be noted that, the trajectory is for the ideal case. If the swimmer whiplashes his arm and feet to create more resistance, then the trajectory won’t be applicable.

Using the equations derived above, the minimum depth of swimming pool required to be manufactured would be determined. The purpose of this exploration is to determine the minimum depth of the swimming pool to ensure no injury of the swimmer due to striking with the floor of pool while diving into the pool. It is obvious that more the velocity while getting into the pool, the diver would go deeper into the pool. Hence, in this exploration, the velocity of the diver while diving from a height of 10 m would be considered to estimate the minimum depth of the swimming pool. To determine the ideal depth of the swimming pool, an angle of 15° will be considered.

Therefore, the required values of the equation (13) are as follows:

*v = *14* m∙s ^{-}*

θ = 15°

*ρ _{b} = 985 kg∙m^{-}*

*ρ _{w} = *1000

Plugging in the above values in equation (13):

\(s_y=\frac{1}{2}\bigg[\frac{(vcoscos\ θ)^2}{\big(\frac{ρ_w}{ρ_b}-1\big)}\bigg]\)

\(s_y=\frac{1}{2}\bigg[\frac{\{\ 14coscos(15°)\}\ ^2}{\big(\frac{1000}{985}-1\big)}\bigg]\)

*s _{y} ≈ *6004

From the above calculation, it is clear that a swimming pool with a depth of 6km (approximately) is not possible. The possible ways to decrease obtaining such a greater depth would be:

- Increase the angle of the motion (feet with respect to vertical axis) with respect to the vertical axis to decrease the velocity of the vertical component of the motion.
- Generation of external resistance by whiplashing arms or feet of the swimmer.

However, it must be noted that the expression of \(s_y=\frac{1}{2}\bigg[\frac{\{\ 14coscos(15°)\}\ ^2}{\big(\frac{1000}{985}-1\big)}\bigg]\) may not yield an accurate value as the values used in this expression are presumed or assumed.

To explore the equation of motion for a swimmer using the parametric equations of motion and thus elucidate the terminal velocity of the swimmer and the minimum depth of the pool for a safe swimming using equation solving:

- The equation of trajectory of the swimmer inside the swimming pool would be: \(y=\frac{1}{2}\bigg(1-\frac{ρ_w}{ρ_b}\bigg)\bigg(\frac{x}{vsinsin\ θ}\bigg)^2\)
*+ x*cot*cot*θ where,*y*is the vertical displacement,*x*is the horizontal displacement, ρ_{w}is the density of water, ρ_{b}is the density of human body, θ is the angle of curvature while getting into the pool with respect to the vertical axis,*v*is the initial velocity of diving at the surface of the water. - The equation of trajectory of the swimmer outside the swimming pool would be:
*y*= 4.9*t*^{2}, where*t*is time. - The trajectory of the swimmer is parabolic in nature. However, it should be noted that, the trajectory is for the ideal case. If the swimmer whiplashes his arm and feet to create more resistance, then the trajectory won’t be applicable.
- The equation of maximum depth attained by the swimmer would be: \(s_y=\frac{1}{2}\bigg[\frac{(vcoscos\ θ)^2}{\big(\frac{ρ_w}{ρ_b}-1\big)}\bigg]\)
- In the case study to obtained to determine the depth reached by the diver without whiplashing and diving from a height of 10 m by making an angle of curvature using his feet right after getting into the pool of 15°, the obtained depth is around 6 km.
- The possible ways to decrease obtaining such a greater depth would be: Increase the angle of the motion (feet with respect to vertical axis) with respect to the vertical axis to decrease the velocity of the vertical component of the motion and generation of external resistance by whiplashing arms or feet of the swimmer.
- The terminal velocity of the diver at the surface of the water of the swimming pool after dividing from a height of 5 m is 9.8 m∙s
^{-1}, 7.5 m is 12.1 m∙s^{-1}and 10 m is 14.0 m∙s^{-1}. - The equation of the terminal velocity of the diver on the surface of the water of the pool is: v = 9.8t where t is the time till which the diver has been freely falling in seconds.

Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student

- The secondary data used in this exploration for case study have been used are collected from authenticated websites. This has extensively decreased the random error in the exploration.
- The motion of the swimmer has been divided into two parts. This is increased the efficiency and accuracy of the trajectory of the swimmer.
- As the value obtained in the case study is very high which is not practically feasible. Hence, possible ways are suggested to decrease the depth obtained by the swimmers.
- One of the motives of the exploration was not served. The minimum depth that one swimming pool should have could not be determined. However, the possible ways to decrease the depth obtained by the diver has been discussed.
- The resistance faced by the diver inside the swimming pool depends on a lot of parameters, such as, material of costume, density of skin hair and many more. However, these parameters were not addressed in this exploration. It should be noted that, in a mathematics internal assessment, it is not possible to determine the resistance offered by material of costume or density of skin hair. The only way to incorporate such parameters is secondary data which is not available in the internet.
- The trajectory of the swimmer is parabolic in nature. However, it should be noted that, the trajectory is for the ideal case. If the swimmer whiplashes his arm and feet to create more resistance, then the trajectory won’t be applicable.

To further extend the discussion, the limitations faced in this exploration should be something that one should aim at. Out of the three mentioned limitations, the most important is the first one. Hence, to eradicate the limitation, the effect of whiplashing of the swimmer in the generation of resistance and hence, trajectory of the swimmer inside the swimming pool would be determined. Consequently, the minimum depth of the swimming pool required for safe diving could be determined. Hence, the research question could be framed as: *To explore the equation of motion for a swimmer using the parametric equations of motion and thus elucidate the minimum depth of the pool for a safe swimming considering the whiplashing of the swimmer.*

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Dr. Adam Nazha

Top IB Math Tutor: 45/45 IBDP, 7/7 Further Math, 7 Yrs Exp, Medicine Student