IB Physics 2.1 (part 1): Motion
Preface
Hello! This is where we’ll actually start our study of Physics. The previous chapter helped us develop the tools we’ll require; this is where we’ll actually begin.
In my very first article on this series, I said that Physics helps explain the intricacies of the universe. Before we get to something so abstract, let’s start with something we can digest. How does the world around us work?
How do cars work? How do lifts work? How do professional cliff jumpers (Yes, it’s an actual profession) determine how fast they have to run so that they can land safely in the water? This is where the study of mechanics comes into play.
To start, we’ll be looking at the Physics behind how things move.
This is called kinematics. By the end of this 2 part article on mechanics, you’ll be able to do cool stuff like determine how fast you have to run so that you can avoid some sharp rocks at the bottom of a cliff and land in water safely.
As always, here are the formulas given for this subtopic. As part of this article, I will also be going into how these equations are derived so that these do not seem arbitrary to you. After all, you only understand Physics if you are able to understand the equations. It is these various equations you will come across that help describe how the universe works. Without further a do, let’s do some Physics!
Distance and displacement
One of the most basic quantities you will come across is displacement. Displacement is the vector form of distance and is usually represented by an s. The reason we use s is because we sometimes refer to displacement as spatial position. Displacement is simply the change in an object’s position with respect to a certain point. Let’s say you were at point A and walked 20 m to the right to point B. This means that you have a displacement of 20 m (Sticking to the convention that rightwards and upwards is our positive vector direction)
Distance on the other hand is a scalar. Distance is kind of hard to define but its essentially the sum of all the magnitudes of displacement an object travels between two points.
Both quantities have units of metres (m).
So we’ve walked from point A to point B which gives us a displacement of 20 m. Now what if we walked from point B to point A? Well in this case, our displacement from B to A would be -20 m. This is because we’re going backwards. Our net displacement would be 20–20=0. Why? Well if we go back to our original position, we don’t really have a net change in our position.
Our distance covered however is simply the sum of the magnitudes of displacement. So our distance d = (20 + 20) m = 40 m
Speed and velocity
It is likely you’re familiar with speed. You’ve likely heard the phrase, “The speed of the car is 60 km/h). Velocity is simply the vector version of speed. We are likely aware that the equation for average speed is total distance travelled/time taken. Hence, the units for average velocity is similarly:
The Δ symbol is greek letter capital Delta and it means “change in”. Hence, velocity is the change in position (Also known as displacement) by change in time. This can be written as:
Where s is displacement and t is time. Velocity also has the same units of speed. The SI units for velocity and speed is m/s or ms^-1
So what is the definition of velocity? We define velocity as the rate of change of displacement. What does this mean?
We can illustrate this by drawing a graph where we put time on the x axis and displacement on the y axis. This is called a displacement-time graph or a position-time graph.
This graph essentially shows how an object’s displacement varies with time.
Now we define velocity as change in displacement over change in time. Essentially, this gives us the slope, or the gradient of this curve.
Hence, the gradient for a displacement-time curve is velocity.
Notice how it is pretty simple to calculate the velocity of this object as the slope is constant throughout. If you were calculate the slope, you would get 2 ms^-1 as the velocity of the object.
Now, what if the slope wasn’t constant throughout?
What if our displacement-time graph looked something like this? Notice how the slope isn’t constant throughout? Well, to find the instantaneous velocity of the object, ie the velocity at a particular point, we would have to use a special tool called calculus. More specifically, we would have to use the derivative. As calculus is not part of the IB course, I won’t be going into it.
Nevertheless, we can basically approximate the object’s velocity at different time values by drawing tangent lines (Lines which touch only point on a graph).
The following shows the tangents at different points of the curve. Notice how when the object is just about to turn back (When the displacement goes from positive to negative), the velocity is 0? This is because when an object has negative velocity, it actually starts travelling in the opposite direction.
Hence, it initially starts travelling in the positive direction, then its velocity becomes 0 and it moves in the negative direction.
You will likely hear the term “Start from rest” very often, this simply means that an object has no initial velocity.
Acceleration
Acceleration is defined as the rate of change of velocity. Essentially, acceleration answers the question: how does your velocity change with respect to time?
Therefore, much like how velocity is displacement by time, acceleration is the change in velocity by the change in time.
Acceleration is a vector (It does not have a scalar equivalent) and has units m/s² or ms^-2.
We can rewrite the equation as
However, the initial velocity is actually represented by the letter u. You will see why this is important soon. Hence, the equation can be written as:
If we were to draw a velocity-time graph. Acceleration would be the gradient. Interestingly enough, the area under your velocity time graph is displacement.
Deceleration is simply defined as negative acceleration.
If an object is travelling at constant velocity, then it has no acceleration. This is because the numerator of the equation becomes 0/t which is just 0.
Time
Time is honestly a mind-boggling concept to wrap your head around. However, for now, time is simply well… time. It is a scalar and cannot be negative. What does -3 s mean? Hence, if you’re ever doing a Physics problem and come across negative time, you’ve gone wrong somewhere.
Constant acceleration
Now that we have a general understanding of the kinematical quantities. How do we actually solve some Physics problems?
Luckily for us, we have a set of equations that help us solve problems if an object undergoes constant acceleration. These equations are known as the SUVAT equations. Each letter stands for each kinematical quantity (S for displacement, U for initial velocity, V for final velocity, A for acceleration, and T for time)
These equations only work if acceleration is constant.
Here is a table that summarises the kinematical quantities.
The SUVAT equations are also the equations given to you on the data booklet. Hence, you don’t have to memorise any of these.
Note that if you are solving a SUVAT question, you actually only need to know three of the five quantities above to solve for one quantity. Hence, if you know an object’s displacement, acceleration, and final velocity, you could calculate the time it took to travel without knowing its initial velocity!
In the next few sections, I will be deriving these equations for you.
Derivation of equation 1: v = u + at
For the purposes of all our derivations, we are going to assume that we are studying a car. We have managed to gather the velocity of the car at different points in time and we have thus been able to plot a velocity-time graph.
As these equations only work for constant acceleration, our velocity time graph can only have a straight line so as to ensure our gradient is constant.
We know that acceleration is defined as the change in velocity over time.
Hence, we can write our acceleration equation like so:
Multiplying both sides by t gives us:
By adding u to both sides, we get:
This is our first equation and it’s probably the simplest one. Notice how at gives us a velocity? This leads us to the relationship that the area under an acceleration time graph is actually velocity.
Derivation of equation 2: s = [(v+u)/2]t
Now, what if we wanted to find the displacement? Well, we know that displacement is the area under a velocity time graph. Hence, let’s say we want to find the displacement at some value t=a
Notice how the area under this graph forms a trapezium? Hence, the displacement would be the area of a trapezium.
The formula for an area of a trapezium is:
Where a and b are the lengths of the parallel sides and h is the height. Hence, our parallel sides are our initial velocity and final velocity at t=a. As we can use this equation for any value of t, we can simply write t instead of h. Hence, we get:
Derivation of equation 3: s = ut + (1/2)at²
We have now established that v=u+at. What if we substitute this into our second equation?
Hence we get:
This simplifies to:
Which then simplifies to:
Derivation of equation 4: v²=u²+2as
Remember our first equation: v=u+at
If we rearrange for t we get:
Plugging this value of t into our second equation we get:
Notice how we essentially have a difference of two squares with (v+u) and (v-u)? Hence, if you were to remember the rule (a+b)(a-b)=a²-b² or just expanded it out, you would get:
Multiply both sides by 2a to get
Hence, adding u² to both sides gets us:
Using the SUVAT equations to solve problems
Now that we’ve actually derived the equations, how do we use it solve a problem?
Let’s say we have the following question:
“A car accelerates from rest to a velocity of 10 ms^-1 under constant acceleration for a time t s. Given that it travelled 20 m during this time. Find the value of t.”
Well, let’s start by listing out all the quantities we know.
The first context clue is that it travelled under constant acceleration. Hence, we can use a SUVAT equation.
We know that it started from rest. Hence, u=0 ms^-1. It’s final velocity is 10 ms^-1 so v=10 ms^-1. It also had a displacement of 20 m. Hence s = 20 m
Therefore we know u, v, and s. The question wants us to find t. As we do not know a, we cannot use any equation with acceleration in it. The only equation which links these quantities, without acceleration, is equation 2.
If we rearrange the equation for t we get:
We can now plug in our values to get:
Hence, the time travelled is 4 seconds.
This is the end of the first part as it’s already become quite long. In the next part, we’ll be finishing up motion by looking at projectile motion and terminal velocity!