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# Tutor profile: Jordan Y.

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Jordan Y.
Maths and Physics tutor for over 4 years, Oxford educated
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A car at rest begins travelling with constant acceleration. It reaches $$15ms^{-1}$$ and begins decelerating. It has a constant deceleration for $$20s$$ before coming to a halt. The car travels $$450m$$ throughout this journey. Find the total time taken, $$t$$, for the car to complete its journey.

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Jordan Y.

To answer this question, let's consider the two parts of the journey separately. For the first part of the journey, we know the initial speed ($$0ms^{-1}$$) and final speed ($$15ms^{-1}$$) of the car. For the second part of the journey, we know the initial speed ($$15ms^{-1}$$) and final speed ($$0ms^{-1}$$) of the car, as well as the fact that it takes $$20s$$ to go from its initial to final speed. We don't have enough information about the first part of the journey to calculate any meaningful values but for the second part, we can use the given values to calculate the distance traveled whilst decelerating by using the formula: $$s = \left(\frac{v + u}{2}\right)t$$ Substituting $$u = 15ms^{-1}$$, $$v = 0ms^{-1}$$, and $$t = 20s$$: $$s = \left(\frac{0 + 15}{2}\right)20 = 150m$$ Knowing this, we can calculate the distance traveled in the first part of the journey: $$450 - 150 = 300m$$ Which we can then use to calculate the time taken for the first part of the journey, again using the formula: $$s = \left(\frac{v + u}{2}\right)t$$ But this time rearranging it to get: $$t = \left(\frac{2s}{v + u}\right)$$ Substituting $$u = 0ms^{-1}$$, $$v = 15ms^{-1}$$, and $$s = 300m$$: $$t = \left(\frac{2(300)}{15 + 0}\right) = 40s$$ Therefore, we can conclude that the first part of the journey took $$40s$$ whilst the second part of the journey took $$20s$$, bringing the total journey time to $$60s$$.

### Subject:Applied Mathematics

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Question:

As you know, when you fold a sheet of A4 paper in half, your resulting sheet of paper will be A5 sized. More importantly than being half the size of A4 paper, A5 paper also has the same ratio between its long edge and short edge as A4 does. What numerical value do you get if you divide the length of the long edge by the length of the short edge on A4 paper?

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Jordan Y.

To answer this question, we can start by drawing out a rectangle (our A4 piece of paper) and labeling the side lengths. Let's give the short edge length $$y$$ and the long edge length $$x$$. Since the question asks for the value of the length of the long edge divided by the short edge, we know that we are trying to find the value of $$\frac{x}{y}$$. Now that we have some variables to work with, we can put the information in the question to use. We know that when folded in half, the sheet of paper retains the ratio between its long and short edges. When we fold the paper in half (to make an A5 sheet), we fold the longer side in half, so the new values for the sides are: long side: $$y$$ short side: $$\frac{x}{2}$$ Now we can see that the ratio between the long and short sides on the smaller sheet is $$\frac{y}{(\frac{x}{2})}$$. This can also be written as $$\frac{2y}{x}$$. From the question, we know that this new expression for the ratio between the long and short sides of the paper must be the same as the ratio on the A4 piece of paper. Therefore we can conclude that: $$\begin{equation*} \frac{2y}{x} = \frac{x}{y} \end{equation*}$$ Rearranging to find $$\frac{x}{y}$$ (as asked to do in the original question): \begin{align*} \frac{x^2}{y^2} &= 2\\\\ \left(\frac{x}{y}\right)^2 &= 2\\\\ \frac{x}{y} &= \sqrt{2} \end{align*} This tells us that the ratio on standardized paper is $$\sqrt{2}$$

### Subject:Calculus

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Question:

Suppose a curve, $$C$$, has equation: $$y = f(x)$$. Given that the equation of the tangent at $$x = a$$ is: $$y = mx + c$$ Find $$f(a)$$ and $$f'(a)$$ in terms of $$c$$, $$m$$, and $$a$$.

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Jordan Y.

This is a nice introductory question to calculus and introduces some important terminology. We must be able to pick out relevant pieces of information and approach the problem with them in mind. To begin, it is best to think about what we are being asked to find. In this case, we are asked for two algebraic values: $$f(a)$$ and $$f'(a)$$. $$f(a)$$ can be described simply as the $$y$$ value of the curve $$C$$ when $$x = a$$. To find this, we must recall that the tangent intersects the curve at a single point. In the question, we are told that the tangent is at $$x = a$$. This means that the $$y$$ value of the tangent is the same as the $$y$$ value of the curve when $$x = a$$. This makes finding $$f(a)$$ a simple exercise in substitution when we consider that the equation for the tangent is $$y = mx + c$$: $$y = ma + c$$ $$y = f(a)$$ $$f(a) = ma + c$$ $$f'(a)$$ can be described as the instantaneous gradient of $$C$$ when $$x = a$$. In order to calculate this, we must recall that the tangent of a curve at some point, $$p$$, will have the same gradient as the the curve at point $$p$$. Because of this, we deduce that $$f'(a) = m$$. So we have: $$f(a) = ma + c$$ $$f'(a) = m$$

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