Ellie S. IB Maths tutor, GCSE Maths tutor, IB Physics tutor, GCSE Phy...

Ellie S.

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Degree: Physics (Bachelors) - Oxford, Lady Margaret Hall University

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About me

About Me:

Hi! I'm Ellie, I'm 19, and I'm a Physics student at Oxford University. My strengths are maths and physics, although I did a number of different subjects in sixth form, so I am also very familiar with Latin, English, and History.

I studied the International Baccalaureate, which means that I am in a good position to tutor IB subjects, including HL Physics, Maths and Further Maths. I also did A-Level Physics alongside my GCSEs, so I am very experienced in this course too (specifically AQA). I achieved 12 A*s in my GCSEs and 44 points in my IB Diploma. 

I hope that I am patient and clear, and that you find the sessions interesting! I love my subject, and I hope that this comes across to you too. 

Tutor Sessions:

I will follow your lead to decide which topics we cover and how to cover them.

The secret to mathematical sciences is practice, so the aim for each topic is to understand the theory - I will try different ways of explaining an idea until we find one that works for you - and then learn to apply this through a cycle of practice questions and feedback. Exam technique can also do a lot for grades in maths and physics.

University Applications:

I would be delighted to help with these too! I am very familiar with the Oxford application system, particularly for Physics, and I will happily give sessions for personal statements, PAT tests, and Oxbridge interviews.

If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session!' We can go through your subject, exam board, target, and areas with which you'd like help.

I look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Physics A Level £20 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Further Mathematics IB £20 /hr
Maths IB £20 /hr
Physics IB £20 /hr
-Personal Statements- Mentoring £20 /hr
.PAT. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
Physics (HL)Baccalaureate7
Maths (HL)Baccalaureate7
Further Maths (HL)Baccalaureate7
Latin (SL)Baccalaureate7
English (SL)Baccalaureate7
History (SL)Baccalaureate7
PhysicsA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

11/04/2016

Currently unavailable: for new students

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Questions Ellie has answered

A cricketer throws a ball vertically upwards so that the ball leaves his hands at a speed of 25 m/s. Calculate the maximum height reached by the ball, the time taken to reach max. height, and the speed of the ball when it is at 50% max. height.

We can see that this is a suvat question, since it involves a projectile. The first step for any suvat question is to list the known suvat variables for the point we're looking at (in this case, the point of maximum height): s = h (we do not yet know the maximum height, h) u = 25 m/s (initial...

We can see that this is a suvat question, since it involves a projectile. The first step for any suvat question is to list the known suvat variables for the point we're looking at (in this case, the point of maximum height):

s = h (we do not yet know the maximum height, h)

u = 25 m/s (initial velocity is given in the question)

v = 0 m/s (velocity at maximum height is zero, because here the ball must change direction from upwards to downwards)

a = -9.8 m/s2 (acceleration is always -9.8 m/s2 for an object free in the air - it is negative because the gravity causing it always acts downwards towards the Earth's centre)

t = ? (we do not yet know the time taken to reach maximum height)

We can then decide which suvat equation to use from which variables are known. In this case, we do not know t, so the best equation to use is v= u2 + 2as, which does not have t in it. Substitute in the numbers:

0 = 252 + 2(-9.8)(h) = 625 - 19.6h

h = 32 m

Now that we know s, u, v and a, we could use any of the equations to find t. We will choose v = u + at for simplicity:

0 = 25 + (-9.8)(t) = 25 - 9.8t

t = 2.6 s

We then repeat the process for the final part of the question:

s = 0.5 x 32 = 16 ; u = 25; v = ?; a = -9.8; t = ?

Again, we do not know t at 50% of the maximum height, so we will use v2 = u2 + 2as:

v2 = 252 + 2(-9.8)(16) = 311

v = 18 m/s

This gives us the answers h = 32 m, t = 2.6 s, v = 18 m/s.

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5 months ago

194 views

Differentiate y = e^(x^2 - 3x).

This question is an example of the chain rule for differentiating.  Firstly, identify the inner function. In this case, it is x2 - 3x. This function must be differentiated first: d/dx (x2 - 3x) = 2x - 3 Secondly, identify the outer function. In this case, it is ez, where z = x2 - 3x. This fu...

This question is an example of the chain rule for differentiating. 

Firstly, identify the inner function. In this case, it is x- 3x. This function must be differentiated first:

d/dx (x2 - 3x) = 2x - 3

Secondly, identify the outer function. In this case, it is ez, where z = x2 - 3x. This function must be differentiated second:

d/dz (ez) = e 

The final differentiated result is the derivative of the inner function multiplied by the derivative of the outer function:

dy/dx = (2x - 3)e= (2x - 3)ex^2 - 3x

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5 months ago

176 views

Find cos4x in terms of cosx.

As with any question involving sines and cosines, consider complex numbers as a likely way to find the answer. For this particular kind of question, where sin/cos of a multiple of x is needed in terms of sinx/cosx, or vice versa, the complex number representation z = cosx + isinx must be expan...

As with any question involving sines and cosines, consider complex numbers as a likely way to find the answer. For this particular kind of question, where sin/cos of a multiple of x is needed in terms of sinx/cosx, or vice versa, the complex number representation z = cosx + isinx must be expanded in two ways. 

Firstly, expand using de Moivre's Theorem:

z = cosx + isinx

z4 = (cosx + isinx)4 =  cos4x + sin4x

Then, expand using the binomial expansion formula to get powers of cosx:

z4 = (cosx + isinx)4 = cos4x + 4icos3xsinx + 6i2cos2xsin2x + 4i3cosxsin3x + i4sin4x

                               = cos4x + 4icos3xsinx - 6cos2xsin2x - 4icosxsin3x + sin4x

Equate the real parts of both expansions to get cos equivalence:

cos4x = cos4x - 6cos2xsin2x + sin4x

Use cos2x + sin2x = 1 as a substitution:

cos4x = cos4x - 6cos2x(1-cos2x) + (1-cos2x)2

          = 8cos4x - 8cos2x + 1

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5 months ago

188 views
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