# A-level Mathematics/MEI/M1
## Forces

Displacement ($m$), velocity ($ms^{-1}$), acceleration ($ms^{-2}$) and force ($N$) are all vectors. Vectors have a direction and a magniture, but are usually defined in form ${\begin{pmatrix}i\\j\end{pmatrix}}$, where i is usually eastwards (x-axis) and j is northwards (y-axis). The absolute value (also known as the magnitude) of this vector is $|r|={\sqrt {i^{2}+j^{2}}}$ and the direction from the positive horizontal i is $\theta =\arctan {\frac {j}{i}}$.
When considering problems with forces, it is usually best to consider the system as a whole and then individual components on their own. You should be familiar with ${\overrightarrow {F}}=m{\overrightarrow {a}}$ (Force = mass times acceleration), in which the 'net' exerted force on an object of mass m causes it to accelerate at acceleration a.

#### Example

A uniform ball of mass 5 kg has force ${\begin{pmatrix}10\\15\end{pmatrix}}N$ acting on it. Calculate the acceleration.
You would solve this by using $a={\frac {F}{m}}$: ${\frac {\begin{pmatrix}10\\15\end{pmatrix}}{5}}={\begin{pmatrix}2\\3\end{pmatrix}}ms^{-2}$ If necessary, you can find the magnitude of this acceleration: $|a|={\sqrt {2^{2}+3^{2}}}={\sqrt {13}}$.

### Force components

It is always the case that an examination paper will ask you to find the appropriate force components, and those will usually involve 'tensions' and 'thrusts' that are caused by strings or rods. You should not worry about inner processes of them and just consider that they have a force all throughout acting on both sides. So if you put two boxes and join when with an inextensible string and pull those boxes apart, tension will build up inside the string. Likewise, for a rigid rod, if you join a car with a caravan and are braking, the rod will experience large compression (thrust) and cause a backwards force against the caravan.
Since forces are vectors (as well as are the others named), if you are given a force in form of magnitude and direction, you can easily calculate the i and j components of the vectors by using trigonometry. The most commonly used trigonometry in this module is the basic $r_{V}=|r|\sin {\theta }$ and $r_{H}=|r|\cos {\theta }$.
Very common problems involve an inclined plane at a certain angle and a couple of forces acting on an object that is located on this inclined plane. You will have to resolve forces parallel and perpendicular to the plane, where the parallel force will always be with $\sin \theta$ and perpendicular (usually reaction) with $\cos \theta$.

## Kinematics

You should be familiar with Suvat equations of motion that describe constant acceleration.

## Calculus in solving problems

Acceleration is the derivative of velocity. Velocity is the derivative of displacement. Likewise, displacement is the integral of velocity and velocity is the integral of acceleration. Displacement is the integral of acceleration. For example, if we consider kinematics, the equation $s=ut+{\frac {1}{2}}at^{2}$ for constant acceleration (with respect to time) would come from: $s=\iint a\,dt\,dt=\int at+u\,dt={\frac {1}{2}}at^{2}+ut+c=c+ut+{\frac {1}{2}}at^{2}$
A very common mistake while integrating is to leave out the constant. Another, and more painful mistake is solving a given integral (say, between t = 3 and t = 6), where you have to find the overall distance travelled and not the displacement (the velocity becomes negative, say, at, t = 5, making displacement less) - so you have to integrate from t = 3 to t = 5 and t = 5 to t = 6, adding both of them as positive values.

## Projectiles

A common problem will be to solve the throwing of a ball over a wall or so. In projectiles, motion has to be considered in j and i directions separately. If you are asked to find the range of a projectile, the best method is to find the time at which the j direction is at 0 (t = 0s, of course, is the first possibility) and then substitute the time back in to find the i direction.