Kinematics is the study of motion in space that is realised in the course of time. The crucial difference between kinematics and dynamics is that kinematics does not consider the physical laws of motion. This may seem strange at first to consider motion apart from its physical laws, but it makes absolute sense. At any rate, we will proceed with a general overview of kinematics and solve some problems related to it.
At any rate, let us begin the study of kinematics. The first place to begin is that, from some simple experiments, it is possible to infer a general rule of motion: objects always free fall at the same rate, independent of their shape, size and mass. The effect of air resistance can mask this fundamental truth by producing a different kind of force, but we will not consider that in this part. We will only consider motions made without air resistance. Under such conditions, a body will fall in a special way. If it falls \(c\) distance in the first unit of time, it will fall \(3c\) distance in the second unit of time, \(5c\) distance in the third unit of time and so on. The consequence of this is that if you take the coefficients and add them up, an interesting pattern emerges. \(1\), \(1+3=4\), \(1+3+5=9\). It forms, in other words, the perfect squares! This is an amazing result if you consider it.
Now, we will definitely want to encode this knowledge into some sort of equation, so if we do that we obtain
\[s = ct^2\]
where \(s\) is the total distance travelled and \(t\) is some time.
The point here is through only elementary experimentation and some simple reasoning, we have been able to identify a physical equation that describes the motion of objects, without us having to worry about the details of how or why gravity causes these objects to move in the first place. This is the core of kinematics.
After that little introduction, let's move onto, derive and understand the classic SUVAT equations. Any physicist will use these pretty much for their entire career. Consider the very naive (but true!) equation.
\[v = \frac{d}{t}\]
where \(v\) is average speed, \(d\) is distance and \(t\) is time. Consider a body moving while accelerating. At time \(t\) it will have position \(s(t)\), and at some later time it will have position \(s(t + \Delta t)\). If we consider this in our naive equation, we can see very easily that a more sophisticated equation for average speed is given by
\[v = \frac{s(t + \Delta t) - s(t)}{\Delta t}\]
Why don't we try plugging in our equation \(s = ct^2\) into what we just derived?
\[s(t + \Delta t) - s(t) = c(t + \Delta t)^2 - ct^2 = 2ct \cdot \Delta t + c \Delta t^2\]
\[v = \frac{2ct \cdot \Delta t + c \Delta t^2}{\Delta t} = 2ct + c \Delta t\]
Now consider that we have a \(\Delta t\) in this equation. What happens when we make this time step smaller and smaller? The two points we are considering become closer and closer together. When the time step becomes vanishingly small, the distance between the two points also becomes vanishingly small. At this point, it no longer makes sense to discuss the "average" speed over some time, since in effect virtually no time has passed. Thus, we are now discussing the instantaneous speed.
\[v = 2ct\]
More formally this is the calculus technique of a limit.
\[\frac{ds}{dt} = \lim_{\Delta t \to 0} \left( \frac{s(t + \Delta t) - s(t)}{\Delta t} \right) = v(t) = 2ct\]
You might think that since these equations have been derived by using \(s = ct^2\) from observing the motion of falling bodies, they only apply to falling bodies. However, this is not true. In fact, these equations hold for any system with a constant acceleration. There's a few ways to think about this. One is that although the equations were formulated by us from our intuition of falling bodies, the equations themselves know nothing about gravity. They only know about constant acceleration. Another way is to make an appeal to symmetry arguments. Although we made reference to objects falling "down", we can rotate the entire universe and have those objects "fall" "left" but nothing would change about the motion of the objects. Less abstractly, objects fall just the same in Britain as they do in Australia.
General SUVAT equations
Now we have the foundations of kinematics, let's derive all the generic SUVAT equations.
An acceleration can be defined as a change in velocity in some change in time, which is
\[a = \frac{v - u}{\Delta t}\]
where \(a\) is acceleration, \(v\) is final velocity, \(u\) is initial velocity.
A simple rearrangement of this gives
\[v = u + a \Delta t\]
In the previous section we used the equation \(s = ct^2\), where we were representing the total distance travelled based on an initial unit of distance. We might not always have the first unit of distance travelled though. Can we make this more useful? Yes, of course! First, consider another way to write this equation. We know from a naive equation that distance = speed \(\times\) time. What is that speed though, if the speed is changing in time? Working that out in detail is hard, but we can simplify this a lot by realising that no matter what complex shape the speed took, the total, final distance will be the average speed \(\times\) time. As an equation
\[d = \bar{v}t\]
where \(\bar{v}\) is the average speed
Why does using the average speed work? Consider the following graph. This is a speed by time graph, which means the area under the lines is the distance travelled in each case. Try calculating the area under the "Linear Acceleration" line and the "Constant Velocity" line to convince yourself the distance travelled is constant.
Since the speed is changing though, what is the average speed? That is a pretty complex question in general, but we can simplify it by limiting the discussion from any arbitrary acceleration to a constant acceleration. In that instance, it is trivial to see that the average speed is given by \(\frac{1}{2} (u + v) \), where again, \(v\) is the final speed and \(u\) is the initial speed. Now we can substitute this into our naive equation and get a new equation for distance.
\[s = \frac{1}{2}(u + v) \Delta t\]
Now that we have \(v = u + a \Delta t\) and \(s = \frac{1}{2}(u + v) \Delta t\) we can try to substitute them into one another
\[\Delta t = \frac{v - u}{a}\]
\[s = \frac{1}{2}(u+v) \cdot \frac{v-u}{a} = \frac{(u+v)(v-u)}{2a}\]
\[2as = (u+v)(v-u) = v^2 - u^2\]
\[v^2 = u^2 + 2as\]
Now, let's try substituting \(v = u + a \Delta t\) directly into \(s = \frac{1}{2}(u + v) \Delta t\)
\[s = \frac{1}{2}(u+u+a \Delta t) \Delta t = \frac{1}{2}(2u \Delta t + a \Delta t^2)\]
\[s = u \Delta t + \frac{1}{2} a \Delta t^2\]
Now we can finally reflect on what \(s = ct^2\) really means. Remember, that equation was for a body which was initially at rest. What we just derived describes the same process but in a much more general form. If \(u = 0\) then \(s = \frac{1}{2} a \Delta t\). Compare the equations and we see that \(c = \frac{1}{2} a\)! This is an amazing result! I don't want to dwell on this so much to become pleonastic, but the distance in unit time we were considering in the first section was exactly equal to half the constant acceleration.
Finally, let's consider substituting \(u = v - a \Delta t\) into \(s = \frac{1}{2}(u + v) \Delta t\). I won't write all the algebra out for this one since it proceeds along parallel lines to the previous derivation (give it a go yourself if you are not convinced!)
\[s = v \Delta t - \frac{1}{2} a \Delta t^2\]
Summary
We can describe any motion that has linear acceleration with the following equations
\[v = u + a \Delta t\]
\[s = \frac{1}{2}(u + v) \Delta t\]
\[v^2 = u^2 + 2as\]
\[s = u \Delta t + \frac{1}{2} a \Delta t^2\]
\[s = v \Delta t - \frac{1}{2} a \Delta t^2\]
Examples
Let's try a few examples to be sure we understand the concepts
Consider dropping a ball from a height of \(100m\), with no starting velocity. How long will it take to first hit the floor? How fast will it be travelling just before it hits the ground?
First, assume the ball is on earth, so the acceleration will be a constant \(9.81 ms^{-2}\) downward.
Let's fill out the variables we know and don't know so far
\[a = 9.81ms^{-2}\]
\[s = 100m\]
\[u = 0ms^{-1}\]
\[v = ?\]
\[t = ?\]
We need to find \(v\) and \(t\). Using \(s = u \Delta t + \frac{1}{2} a \Delta t^2\) to find the time. Since \(u = 0ms^{-1}\)
\[s = \frac{1}{2} a \Delta t^2\]
A simple rearrangement gives
\[\Delta t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{200}{9.81}} = 4.51s\]
We can now use \(v^2 = u^2 + 2as\) to find the final velocity
\[v = \sqrt{2as} = \sqrt{2 \times 9.81 \times 100} = \sqrt{1962} = 44.29ms^{-1}\]
Now for another problem. Two cars are driving, with one starting \(500m\) behind the other. Both cars are travelling at \(27ms^{-1}\). On an infinite straight road, how much linear acceleration must the rear car experience to overtake the front car in 30 seconds? How fast will it be travelling when it does?
Now, this problem is a little trickier. The front car experiences no acceleration, so calculating its distance travelled in 30 seconds is trivial - \(s = vt = 27 \times 30 = 810m\). The second car must travel this distance plus 500 metres in 30 seconds. We can use \(s = u \Delta t + \frac{1}{2} a \Delta t^2\) again.
\[s - u \Delta t = \frac{1}{2} a \Delta t^2\]
\[a = \frac{2}{\Delta t^2}(s - u \Delta t) = \frac{2}{30^2}(1310 - 27 \times 30) = 1.11ms^{-2}\]
We can use \(v = u + a \Delta t\) to calculate the final velocity now simply as \(v = 27 + 1.11 \times 30 = 60.3ms^{-1}\). That's about 135 miles per hour!
Projectiles and Vectors
Up until now we have considered motion only in one direction. What about two dimension motion? Let's consider a person throwing a ball at an angle. As you probably know, balls thrown like this form arcs in the air. They also travel up and down and also forward. Let's first consider a totally generalised problem.
A ball is thrown with some velocity. How far will it travel, how high will it get, how long will it be in the air, and what is its final velocity?
Since we want to keep this totally general, let's use vectors and write out what we know
\[\vec{u} = \colvec{u_x\\u_y}\]
\[\vec{a} = \colvec{0\\-g}\]
\[\vec{v} = \colvec{?\\?}\]
\[\vec{s} = \colvec{?\\?}\]
\[\Delta t = ?\]
The arrow on top of letters like this \(\vec{a}\) indicates a vector - this is a collection of values which describe a space. In this instance, a two dimensional space needs two values, hence the vectors \(\colvec{u_x\\u_y}\). The convention I will follow is the top value represents the left-right direction, or \(x\) direction and the bottom value represents the up-down direction or \(y\) direction. I have chosen to define right and up as the positive direction. You can choose any axis you like, as long as you stay consistent.
There's a lot going on now, so the easiest place to start is to get the time of flight. The vectors themselves are suggestive of an important point to note - the \(x\) and \(y\) directions are independent. In other words, we can consider the up-down motion totally independently of the left-right motion. Let's consider the time of flight. The ball will stop when it hits the ground, and we can reason that we can obtain this time as the time it takes for the ball to rise to its maximum height and fall to the ground again.
Therefore, we can use \(v = u + a \Delta t\) with setting \(v = 0\) to get the time to reach the highest point. We must be careful to take the variables happening in the \(y\) direction only.
\[0 = u_y + a \Delta t\]
\[\frac{-u_y}{a} = \Delta t = \frac{-u_y}{-g} = \Delta t = \frac{u_y}{g}\]
That is the time to reach the highest point, the time to fall again will be exactly equal. In other words, the total time of flight is \(\frac{2u_y}{g}\). This is an important result, as it lets us calculate everything else. Now is a good time for us to really use the vectors. Let's find the distances travelled. If we use the equation \(s = u \Delta t + \frac{1}{2}a \Delta t^2\) we can find the distances.
\[\colvec{s_x\\s_y} = \colvec{u_x\\u_y} \frac{u_y}{g} + \frac{1}{2} \colvec{0\\-g} \left( \frac{2u_y}{g} \right)^2\]
Notice how I have written out the full vectors and placed them into the equation. The same physical law applies to both the \(x\) and \(y\) direction, only the numbers change. We don't need to do anything but read off the values of \(s_x\) and \(s_y\).
\[s_x = \frac{2u_x u_y}{g}\]
\[s_y = \frac{2u_y^2}{g} - \frac{1}{2} g \frac{4u_y^2}{g^2} = \frac{2u_y^2}{g} - \frac{2u_y^2}{g} = 0\]
First, don't get confused that the \(u_y\) and \(-g\) are on the bottom - these are not fractions! These still multiply when you convert them into equations. Second, the 0 result for \(s_y\) might surprise you, after all the ball will move up and down. However, you need to consider the difference between distance and displacement. Up until now, we've been a little loose about them, but they are subtly different. Distance is how far something travels in total. Displacement is how far something travels from the starting location. Although the ball will be in physically in a different location when it lands, from the perspective of the \(y\) direction it will be at the same height. Hence, over the whole operation there is no displacement in the \(y\) direction.
We still want to find the maximum height of the projectile though, and we can use \(s = u \Delta t + \frac{1}{2}a \Delta t^2\) but this time with \(\Delta t = \frac{u_y}{g}\).
\[s_y^\prime = \frac{u_y^2}{g} - \frac{1}{2} g \frac{u_y^2}{g^2} = \frac{u_y^2}{g} - \frac{1}{2} \frac{u_y^2}{g} = \frac{1}{2} \frac{u_y^2}{g}\]
I use \(s_y^\prime\) to indicate this maximum height.
All we need to find now is the final velocity of the ball. We can use \(v^2 = u^2 + 2as\) to find this.
\[\colvec{v_x^2\\v_y^2} = \colvec{u_x^2\\u_y^2} + 2 \colvec{0\\-g} \colvec{s_x\\s_y}\]
We already have an expression for \(s_x\) but I won't substitute that till the end for clarity. Let's place 0 in for \(s_y\) though.
\[\colvec{v_x^2\\v_y^2} = \colvec{u_x^2\\u_y^2} + 2 \colvec{0\\-g} \colvec{s_x\\s_y}\]
Solving for \(v_x\) first:
\[v_x^2 = u_x^2\]
\[v_x = \pm u_x\]
The 0 in the second term cancels it out
Now solving for \(v_y\):
\[v_y^2 = u_y^2\]
\[v_y = \pm u_y\]
There's an amazing symmetry to this problem! The final an initial velocities are the same in magnitude to the start velocities. We need to sort those plus-minus signs out though. Consider the \(x\) direction first. The ball is thrown in the positive direction and there's no reason to think that it changed direction. In fact, it will still be moving at exactly the same speed in this direction. So the solution here is \(v_x = u_x\). Now, consider the \(y\) direction. There are two places where \(s_y=0\), at the very beginning (where the positive \(v_y = u_y\)) is satisfied. The other is right at the end, where the \(v_y = -u_y\) is relevant - just think, it will be falling down at this point.
We have now solved the many aspects of this problem, to summarise:
\[\vec{u} = \colvec{u_x\\u_y}\]
\[\vec{a} = \colvec{0\\-g}\]
\[\vec{v} = \colvec{u_x\\-u_y}\]
\[\vec{s} = \colvec{\frac{2 u_x u_y}{g}\\0}\]
\[\Delta t = \frac{2u_y}{g}\]
\[s_y^\prime = \frac{1}{2} \frac{u_y^2}{g}\]
Let's plug in some numbers. For this example, let's say that \(u_x = u_y = 10ms^{-1}\). In other words, the ball is thrown at a 45° angle at a total velocity of \(14.14ms^{-1}\). We will of course use \(g=9.81ms^{-2}\).
\[\vec{u} = \colvec{10ms^{-1}\\10ms^{-1}}\]
\[\vec{a} = \colvec{0ms^{-2}\\-9.81ms^{-2}}\]
\[\vec{v} = \colvec{10ms^{-1}\\-10ms^{-1}}\]
\[\vec{s} = \colvec{\frac{2 \times 10 \times 10}{9.81}\\0} = \colvec{20.39m\\0m}\]
\[\Delta t = \frac{2u_y}{g} = \frac{2 \times 10}{9.81} = 2.04s\]
\[s_y^\prime = \frac{1}{2} \frac{10^2}{9.81} = 5.10m\]
Below you can see a plot of this:
Try reading key places off the graph to confirm it works