**About me**

Hi, I'm Tim, a 3rd year undergraduate Computer Scientist at the University of York. After my degree I hope to do a Masters in Artificial Intelligence.

I love problem solving and programming, and spend a lot of my time working on my own projects (apps, websites and games). I have **strong conceptual and practical maths skills**, and I enjoy finding ways to put them to more **interesting uses** (such as creating sensors for raspberry pi's!).

I enjoy creative outlets, particularly drumming. I play in a alternative-rock band with a group of friends from my University. On Tuesdays, I captain a Chess team for the University in weekly matches.

**Why I Tutor**

The main attraction of tutoring is that it is an opportunity to **break ideas down** into their core elements and help others understand how those elements **add up to the bigger picture**. When I was studying for my AS-Levels I worked as part of a Young Enterprise company doing exactly that - we programmed an App that breaks down Maths and Physics GCSE questions and solves them step-by-step so that the Student can understand where they have gone wrong!

__The Session__

In my tutoring sessions, **you will lead what we do**.

If there is a problem that you are struggling with, we will break it down until you can explain it back to me **with confidence**. To do this, we will use **analogies**, **step-by-step** **break downs** of Maths and **diagrams **to help you find a way to understanding.

I will make the sessions as **fun** and as **interesting** as possible and try to offer **new ways of thinking** about things that you won't have heard before!

__Applying for Computer Science / Software Engineering / Computing at University?__

While Universities often claim they do not require any previous programming experience, it's great to have on your application because it demonstrates **enthusiasm **and **technical ability**.

If you are thinking about applying for Computer Science or Software Engineering then consider hiring me to tutor you programming.

We can focus on **any idea or project** (Websites/Apps/Games/Software) or **any language** that you want and hopefully you'll get **Key Skills **and you'll have made **something cool** to show off to people!

__What now?__

If you want to find out more then send me a WebMail or click the 'Meet the Tutor Session' button!.

Thanks for reading!

**About me**

Hi, I'm Tim, a 3rd year undergraduate Computer Scientist at the University of York. After my degree I hope to do a Masters in Artificial Intelligence.

I love problem solving and programming, and spend a lot of my time working on my own projects (apps, websites and games). I have **strong conceptual and practical maths skills**, and I enjoy finding ways to put them to more **interesting uses** (such as creating sensors for raspberry pi's!).

I enjoy creative outlets, particularly drumming. I play in a alternative-rock band with a group of friends from my University. On Tuesdays, I captain a Chess team for the University in weekly matches.

**Why I Tutor**

The main attraction of tutoring is that it is an opportunity to **break ideas down** into their core elements and help others understand how those elements **add up to the bigger picture**. When I was studying for my AS-Levels I worked as part of a Young Enterprise company doing exactly that - we programmed an App that breaks down Maths and Physics GCSE questions and solves them step-by-step so that the Student can understand where they have gone wrong!

__The Session__

In my tutoring sessions, **you will lead what we do**.

If there is a problem that you are struggling with, we will break it down until you can explain it back to me **with confidence**. To do this, we will use **analogies**, **step-by-step** **break downs** of Maths and **diagrams **to help you find a way to understanding.

I will make the sessions as **fun** and as **interesting** as possible and try to offer **new ways of thinking** about things that you won't have heard before!

__Applying for Computer Science / Software Engineering / Computing at University?__

While Universities often claim they do not require any previous programming experience, it's great to have on your application because it demonstrates **enthusiasm **and **technical ability**.

If you are thinking about applying for Computer Science or Software Engineering then consider hiring me to tutor you programming.

We can focus on **any idea or project** (Websites/Apps/Games/Software) or **any language** that you want and hopefully you'll get **Key Skills **and you'll have made **something cool** to show off to people!

__What now?__

If you want to find out more then send me a WebMail or click the 'Meet the Tutor Session' button!.

Thanks for reading!

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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4.5from 2 customer reviews

Sue (Parent from Droitwich)

May 4 2016

Excellent as always - thanks Tim

Sue (Parent from Droitwich)

February 17 2016

Knows the subject (maths) very well and is good at getting his point across easily and understandably, was able to help me understand every question i threw at him without a problem, he is personable and easy to talk to. Only issue we had was technical.

**A bullet is fired horizontally from a rifle 1.5m from the ground at 430m/s. How far does it travel and for how long does it travel before it hits the ground?**

*Assuming that air resistance is negligable*

What's important to realize about this question is that **the horizontal velocity of the bullet makes no difference to how long it takes to fall to the the ground.**

Because the gun has not applied any vertical forces to the bullet, the only force affecting the bullet is gravity. This means the bullet takes just as long to fall to the ground as it would if it were dropped, despite it now travelling a large horizontal distance in the duration.

To find the travel time before hitting the ground we have 3 values:

-The displacement from the ground that the bullet must travel, **s = 1.5m**

-The acceleration the bullet experiences. As gravity is accelerating the bullet downwards, **a = g = ~9.81m/s^2**

-The initial velocity of the bullet vertically. As the bullet is stationary vertically (it is only travelling horizontally at the start), **u = 0m**

**We examine our equations of motion, commonly known as SUVAT equations. You may need to learn these for your exam, but some examination boards provide them.**

Because we have **s**, **u** and **a**, and we are looking for the time **t**, the relevant equation is

**s = ut + 0.5(at^2) **

Filling in our values we have:

1.5 = 0t + 0.5(9.81 x t^2)

1.5 = 4.905 x t^2

*Divide 1.5 by 4.905 to find t^2*

t^2 = 0.3058...

We simply find the square root of t^2 to find **t**, the time taken for the bullet to reach the ground:

**t = 0.553s** *(3 significant figures)*

To find the horizontal distance, **d**, that the bullet has travelled before it has hit the ground we can use the equation linking displacement **s **with some velocity **v **over some time duration **t:**

**s = vt**

The horizontal velocity of the bullet, **v = 430**

The time before the bullet hits the ground, **t = 0.553**

So **d = vt = 430 * 0.553 = 238m ***(3 significant figures)*

**A bullet is fired horizontally from a rifle 1.5m from the ground at 430m/s. How far does it travel and for how long does it travel before it hits the ground?**

*Assuming that air resistance is negligable*

What's important to realize about this question is that **the horizontal velocity of the bullet makes no difference to how long it takes to fall to the the ground.**

Because the gun has not applied any vertical forces to the bullet, the only force affecting the bullet is gravity. This means the bullet takes just as long to fall to the ground as it would if it were dropped, despite it now travelling a large horizontal distance in the duration.

To find the travel time before hitting the ground we have 3 values:

-The displacement from the ground that the bullet must travel, **s = 1.5m**

-The acceleration the bullet experiences. As gravity is accelerating the bullet downwards, **a = g = ~9.81m/s^2**

-The initial velocity of the bullet vertically. As the bullet is stationary vertically (it is only travelling horizontally at the start), **u = 0m**

**We examine our equations of motion, commonly known as SUVAT equations. You may need to learn these for your exam, but some examination boards provide them.**

Because we have **s**, **u** and **a**, and we are looking for the time **t**, the relevant equation is

**s = ut + 0.5(at^2) **

Filling in our values we have:

1.5 = 0t + 0.5(9.81 x t^2)

1.5 = 4.905 x t^2

*Divide 1.5 by 4.905 to find t^2*

t^2 = 0.3058...

We simply find the square root of t^2 to find **t**, the time taken for the bullet to reach the ground:

**t = 0.553s** *(3 significant figures)*

To find the horizontal distance, **d**, that the bullet has travelled before it has hit the ground we can use the equation linking displacement **s **with some velocity **v **over some time duration **t:**

**s = vt**

The horizontal velocity of the bullet, **v = 430**

The time before the bullet hits the ground, **t = 0.553**

So **d = vt = 430 * 0.553 = 238m ***(3 significant figures)*

**What is the intergral of 6.x^2 + 2/x^2 + 5 with respect to x?**

To answer a question like this you would **intergrate each phrase** of the addition in turn with** respect to x**.

To intergrate a single *c.x^n* phrase with respect to *x*;

-divide *c* by *(n + 1)*

-raise *n* by *1*

Giving you *(c/(n+1)).x^(n+1).*

**For example the first phrase:** *6x^2;*

*c* is *6*, *n* is *2*;

so the intergral of *6.x^**2* with respect to *x* is* (6/(2 + 1)).x^(2+1) = 2.x^3.*

**For the second phrase,** *2x^-2* which is equivalent to *2/x^2*,

*c* is *2*, and n is *-2*;

so the intergral of the phrase with respect to *x* is *(2/(-2+1)).x^(-2+1) = -2.x^-1. *

**For the final phrase,** *5*, which is in the form of simply '*c*', intergrates easily;

*c* is *5*, n is *0*;

So the intergral of this phrase with respect to *x* is *(5/(0+1)).x^(0+1) = 5.x^1 = 5.x.*

This can be done quickly by thinking of the intergral of a constant *c* as simply *c.x.*

The full intergral of the entire question *6.x^2 + 2.x^-2 + 5* with respect to *x* is:

*2.x^3 -2.x^-1 + 5.x + c*

You may notice the extra *c* added on the end. The *c* occurs because the differentation of any constant *c* is *0*, so there may be any value of *c* in the intergral. This c is called the arbitrary **constant of intergration**, whose value can be found with additional information, producing a **particular intergral**. The important thing to remember though is that when you intergrate a function, **include + c on the end!**

Note: I can't seem to get superscript to work so I'm using ^ to represent 'power of' where *x^y* is *x* to the power of *y.*

**What is the intergral of 6.x^2 + 2/x^2 + 5 with respect to x?**

To answer a question like this you would **intergrate each phrase** of the addition in turn with** respect to x**.

To intergrate a single *c.x^n* phrase with respect to *x*;

-divide *c* by *(n + 1)*

-raise *n* by *1*

Giving you *(c/(n+1)).x^(n+1).*

**For example the first phrase:** *6x^2;*

*c* is *6*, *n* is *2*;

so the intergral of *6.x^**2* with respect to *x* is* (6/(2 + 1)).x^(2+1) = 2.x^3.*

**For the second phrase,** *2x^-2* which is equivalent to *2/x^2*,

*c* is *2*, and n is *-2*;

so the intergral of the phrase with respect to *x* is *(2/(-2+1)).x^(-2+1) = -2.x^-1. *

**For the final phrase,** *5*, which is in the form of simply '*c*', intergrates easily;

*c* is *5*, n is *0*;

So the intergral of this phrase with respect to *x* is *(5/(0+1)).x^(0+1) = 5.x^1 = 5.x.*

This can be done quickly by thinking of the intergral of a constant *c* as simply *c.x.*

The full intergral of the entire question *6.x^2 + 2.x^-2 + 5* with respect to *x* is:

*2.x^3 -2.x^-1 + 5.x + c*

You may notice the extra *c* added on the end. The *c* occurs because the differentation of any constant *c* is *0*, so there may be any value of *c* in the intergral. This c is called the arbitrary **constant of intergration**, whose value can be found with additional information, producing a **particular intergral**. The important thing to remember though is that when you intergrate a function, **include + c on the end!**

Note: I can't seem to get superscript to work so I'm using ^ to represent 'power of' where *x^y* is *x* to the power of *y.*