# A-level Mathematics/OCR/M1/Kinematics of Motion in a Straight Line
## Introduction to Kinematics

Kinematics is the study of the motion of a body, but not the ways the motion is produced. To study a ball's motion through a parabolic path is kinematics. Studying how the ball was thrown in the first place is mechanics. Kinematics is one of the most useful parts of physics, and it has wide range of applications. In fact, a lot of dynamics uses kinematical relations.
When we consider the speed of a body, we do not discuss the direction. A ball thrown up with a speed of 20 m/s is equivalent to a ball thrown in any other direction with the same speed. However, these motions are vastly different and gravity will act differently on the ball. So a complete discussion of the motion of a body depends on the direction as well as the speed. Together we get a vector quantity that is called velocity, and is denoted by ${\vec {v}}$.
then, explain their equivalents: velocity and displacement (use diagrams), and show they are vector quantities (provide example in 1 dimension).

## Constant Acceleration

Explain constant acceleration and why we simplify problems by assuming it is constant, and show how to solve some problems involving motion with a constant acceleration
An object under constant acceleration in 1 dimension has its kinematic quantities governed by the SUVAT equations. S is displacement, U is starting velocity, V is final velocity, A is acceleration, and T is time. Knowing three of these quantities allows the other two to be found. The equations are as follows:
$v=u+at$
$s=ut+{\frac {1}{2}}a{t}^{2}$
$s=vt-{\frac {1}{2}}a{t}^{2}$
$s={\frac {u+v}{2}}t$
${v}^{2}={u}^{2}+2as$
For example, a particle dropped from a ledge 2 m high accelerating under gravity has $u=0{\text{ ms}}^{-1}$, $a=-9.8{\text{ ms}}^{-2}$, and $s=2{\text{ m}}$. If you wanted to find the time it takes to fall, you would substitute these values into $s=ut+{\frac {1}{2}}a{t}^{2}$ and solve for t.

## The Relationship Between Kinematic Quantities

show (t,x) and (t,v) graphs and emphasise memorizing features of each graph:
(i) the area under a (t,v) graph represents displacement,
(ii) the gradient of a (t, x) graph represents velocity,
(iii) the gradient of a (t,v) graph represents acceleration;

## Deriving Kinematic Quantities Using Calculus

We have seen how the (x,t) graph shows the position x at any moment t. This means that the position x is a function of the time t.

### Position

As motion happens to happen in time, the position of a moving point is a function of time. The motion of a point moving along the x-axis may be described by the position x(t) as a function of the time (moment) t. A point moving with constant speed in the positive direction may be described by:
    $x(t)=1+3t$.
From this equation we learn:
    $x(0)=1+3\times 0=1$
which means that the motion started at time t = 0 in the point x = 1. At the moment t=1, the point has reached the point x = 4, as we can see by calculating:
    $x(1)=1+3\times 1=4$.
At the moment t=2, the point has reached the point x = 7:
    $x(2)=1+3\times 2=7$.
Notice that during the first unit of time: from t = 0 to t = 1, the point has moved from x = 1 to x = 4, hence advanced 3 units in distance; during the second unit of time, from t = 1 to t = 2, it moved from x = 4 to x = 7, again advancing 3 units. It seems to advance 3 units in distance every unit of time, meaning it moves with constant velocity of 3.

### Velocity

If the point moves with constant speed t, this speed may be calculated from the time it took to move a certain distance from one position to another. If, as in the example above it took 1 time unit to move a distance 3, i.e. from x = 1 to x = 4, the velocity must be v = 3/1 = 3. We could calculate this velocity from the fact that it took 2 units of time to move a distance 6, namely from x = 1 to x = 7, hence v = 6/2 = 3.
In general we calculate the velocity of a motion with constant speed by looking at the time interval from t = t$_1$ to t = t$_2$, in which the point has moved from has travelled from x = x(t$_1$) to d = x(t$_2$), of as the quotient of the travelled distance and the time it took:
    $v={\frac {\mathrm {travelled\;distance} }{\mathrm {required\;time} }}={\frac {x(t_{2})-x(t_{1})}{t_{2}-t_{1}}}$.
We verify that in the above example the speed is indeed constant 3, by calculating for an arbitrary time interval the speed v:
    $v={\frac {x(t_{2})-x(t_{1})}{t_{2}-t_{1}}}={\frac {(1+3t_{2})-(1+3t_{1})}{t_{2}-t_{1}}}={\frac {3t_{2}-3t_{1}}{t_{2}-t_{1}}}={\frac {3(t_{2}-t_{1})}{t_{2}-t_{1}}}=3$.
For a motion with constant speed v the equation of motion is:
    $x(t)=x(0)+vt$.
In non-general the motion will not be with constant velocity. Let us take an example with:
    $x(t)=1+3t+2t^{2}$.
It starts at t = 0 in x(0) = 1. At the moment t = 1 it reaches x(1) = 1 + 3 + 2 = 6. The average speed during this time interval, or on this traject is:
    ${\overline {v}}={\frac {x(1)-x(0)}{1-0}}={\frac {6-1}{1}}=5$.
But during the final time unit:
    ${\overline {v}}={\frac {x(2)-x(1)}{2-1}}={\frac {(1+6+8)-(1+3+2)}{1}}=9$.
The speed has increased. The speed is also a function v(t) of the time. But what would be the (instantaneous) velocity v(t) at the moment t We may calculate this velocity by taking a (very) small time interval from t to t + h. Then the average speed during this interval is approximately equal to the velocity at the moment t. We get the exact value of the velocity by letting the time interval shrink to zero and taking the limit, which is not just the derivative of the position x(t):
    $v(t)=\lim _{h\to 0}{\frac {x(t+h)-x(t)}{h}}={\frac {d}{dt}}x(t)$.
here is where the original author can put his or her work back in to the page!
just remember to keep it simple though, perhaps no need to explain it using first principles. Just a simple explanation of how to go from one to another (displacement, velocity, acceleration) and back again by differentiation or integration