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Degree: Computer Science (Bachelors)  York University
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.Â
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 alternativerock band with a group of friends from my University. On Tuesdays, I captain a Chess team for the University in weekly matches.â€‹
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.Â When I was studying for my ASLevels 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 stepbystep so that the Student can understand where they have gone wrong!Â
The Session
In my tutoring sessions, you will lead what we do.Â
If there is a problem that you are struggling with,Â we will break it down until you can explain it back to meÂ with confidence. To do this, we will useÂ analogies,Â stepbystepÂ break downs of Maths andÂ diagramsÂ 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 demonstratesÂ enthusiasmÂ andÂ technical ability.Â
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 getÂ Key SkillsÂ 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!.Â
Thanks for reading!
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Sue (Parent) May 4 2016
Sue (Parent) February 17 2016
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.Â
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,Â 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),Â 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.Â
Because we have s, u and a, and we are looking for the timeÂ t, the relevant equation is
s = ut + 0.5(at^2)Â
Filling in our values we have:
1.5 = 0t + 0.5(9.81 xÂ t^2)
1.5 = 4.905 x t^2Â
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:
Â
t = 0.553s (3 significant figures)
To find the horizontal distance, d,Â that the bullet has travelled before it has hit the ground we can use the equation linking displacementÂ sÂ with some velocityÂ vÂ over some time durationÂ t:
s = vt
The horizontal velocity of the bullet,Â v = 430
The time before the bullet hits the ground,Â t = 0.553
SoÂ d = vt = 430 * 0.553 = 238mÂ (3 significant figures)
see moreWhat is the intergral ofÂ 6.x^2Â + 2/x^2Â + 5Â with respect toÂ x?
To answer a question like this you wouldÂ intergrate each phraseÂ of the addition in turn withÂ respect toÂ x.
To intergrate a singleÂ c.x^nÂ phrase with respect toÂ x;
divideÂ cÂ byÂ (n + 1)
raiseÂ nÂ byÂ 1
Giving youÂ (c/(n+1)).x^(n+1).
For example the first phrase:Â 6x^2;
cÂ isÂ 6,Â nÂ isÂ 2;
so the intergral ofÂ 6.x^2Â with respect toÂ xÂ isÂ (6/(2 + 1)).x^(2+1) = 2.x^3.
ForÂ the second phrase,Â 2x^2Â which is equivalent toÂ 2/x^2,
cÂ isÂ 2, and n isÂ 2;Â
so the intergral of the phrase with respect toÂ xÂ isÂ (2/(2+1)).x^(2+1) = 2.x^1.Â
For the final phrase,Â 5, which is in the form of simply 'c', intergrates easily;
Â cÂ isÂ 5,Â n isÂ 0;
So the intergral of this phrase with respect toÂ xÂ isÂ (5/(0+1)).x^(0+1) = 5.x^1Â = 5.x.
This can be done quickly by thinking of the intergral of a constantÂ cÂ as simplyÂ c.x.
The full intergral of the entire questionÂ 6.x^2Â + 2.x^2Â + 5Â with respect toÂ xÂ is:
2.x^3Â 2.x^1Â + 5.x + c
You may notice the extraÂ cÂ added on the end. TheÂ cÂ occurs because the differentation of any constantÂ cÂ isÂ 0, so there may be any value ofÂ cÂ in the intergral. This c is called the arbitraryÂ constant of intergration, whose value can be found with additional information, producing aÂ particular intergral. The important thing to remember though is that when you intergrate a function,Â includeÂ + cÂ 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.
see more