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JCOJCHLCOLCH

Tools (click to expand)

Abstract

Learning Outcomes

  • Define displacement.
  • Define velocity.
  • Define acceleration.
  • Measure velocity and acceleration.
  • Draw and interpret distance-time and velocity-time graphs.
  • Solve problems using the equations of motion.
  • Derive the equations of motion.
  • Solve problems about motion in sports.
  • Measure the acceleration due to gravity

Keywords

  • Distance, displacement
  • Scalar, vector
  • Magnitude, direction
  • Speed, velocity
  • Time
  • Acceleration

Extra Credit: Motion (click to expand)

Distance and displacement

Distance measures how far apart two points are. Depending on exactly what is being measured, this may or may not be a straight line. For example we might measure the length of a table using a straight line, but if we talked about the distance from Dublin to Cork, we would probably take into account the curvy roads that make up the journey. Distance is an example of a scalar quantity: it only has a size, or a magnitude. We would say that the distance between Galway and Dublin is about the same as the distance between Galway and Cork \(\left(\approx 200km\right)\). It doesn't matter where in the country those places are, what matters is the length of road between them.

In physics, we often care about the direction of a quantity if there is one. Those journeys from Galway would be considered different because you need to travel in a different direction in each case. We would often see displacement as a more important quantity:

Displacement is distance in a given direction.

  • Displacement is a vector quantity because it has both magnitude and direction.
  • The unit for displacement is metres \(\left(m\right)\)
  • We usually use \(s\) as the symbol for displacement, since \(d\) is already taken by distance. Think of it as short for space.

When we talk about displacement, we make two distinctions from distance.

  1. Displacement refers only to the straight line distance between two points. Every path taken is simplified to be a straight line path.
  2. The direction of motion is important.
The displacement represents a straight-line distance 20° south of east.

In this section, we consider only linear motion, so there can only be two possible directions of motion in any given question (e.g. forward/back, up/down, East/West etc.). We will look at more complicated motion in later sections.

Speed and Velocity

Speed is a measure of how quickly something is moving and you will remember from JC maths that it is defined as:

\[\text{average speed} = \frac{\text{distance}}{\text{time}}\]

Speed depends on distance, so it is a scalar quantity as well. Direction does not matter when we dicuss speed. If we care about direction, then we must refer instead to velocity:

Velocity is the rate of change of displacement with respect to time.
or \[\text{average velocity} = \frac{\text{displacement}}{\text{time}} \Leftrightarrow v_{avg}=\frac{s}{t}\]

  • Velocity is a vector quantity since direction is important.
  • The unit for velocity is metres per second \(\left(m/s \text{ or }m\cdot s^{-1}\right)\)
  • We usually use \(v\) or \(u\) as the symbols for velocity.
  • Examples
  • Easy Sample Problems
  • Medium Sample Problems
  • Hard Sample Problems

Acceleration

Acceleration is a measure of how how quickly a body's velocity is changing. Since we know velocity has both a magnitude and a direction, there is acceleration if the body is changing its speed and/or its direction. We define it as:

Acceleration is the rate of change of velocity with respect to time.
or \[\text{average acceleration} = \frac{\text{change in velocity}}{\text{time taken}} \Leftrightarrow a = \frac{v-u}{t} \]

  • Acceleration is a vector quantity since direction is important.
  • The unit for acceleration is metres per second squared \(\left(m / s^2 \text{ or } m \cdot s^{-2} \right)\).
  • \(a\) is the usual symbol for acceleration.

When a body is accelerating, velocity can by definition no longer be constant. Our previous equation \(\left(v_{avg}=\frac{s}{t}\right)\) can tell us about average velocity, but not about instantaneous velocity. In particular, we often want to know about the velocity and the beginning and end of a journey. For instance, Usain Bolt's average speed of \(10.44 m/s\) is impressive enough, but consider that he started off still and had to accelerate. In fact his highest velocity has been measured to be \(12.27 m/s\)!

To deal with this, we need to introduce some new equations that account for changing velocity, provided the acceleration is constant.

At higher level, you need to be able to derive these three equations of motion:

To prove: \(v = u + at\)

\[\begin{align*} a &= \frac{v-u}{t} & \text{by definition} \\ \Rightarrow at &= v-u & \\ \Rightarrow v &= u + at & \\ \end{align*}\]

To prove: \(s=ut+\frac{1}{2}at^2\)

\[\begin{align*} v_{avg} &= \frac{s}{t} & \text{by definition} \\ v_{avg} &= \frac{v + u}{2} & \\ \Rightarrow \frac{s}{t} &= \frac{v + u}{2} & \text{combining these two} \\ \Rightarrow s &= \frac{v + u}{2}t & \text{multiplying by }t \\ \Rightarrow s &= \frac{(u + at) + u}{2}t & \text{using }v = u+at \\ \Rightarrow s &= \frac{2u + at}{2}t \\ \Rightarrow s &= ut + \frac{1}{2}at^2 \end{align*}\]

To prove: \(v^2 = u^2 + 2as\)

\[\begin{align*} v &= u + at & \\ \Rightarrow v^2 &= (u + at)^2 & \text{square both sides} \\ \Rightarrow v^2 &= u^2 + 2aut + a^2t^2 & \\ \Rightarrow v^2 &= u^2 + 2a\left(ut + \frac{1}{2}at^2\right) & \text{factoring out }2a \\ \Rightarrow v^2 &= u^2 + 2as & \text{using }s = ut + \frac{1}{2}at^2 \\ \end{align*}\]

These equations are used to solve most problems in linear motion when acceleration is present.

Last modified: 2017-12-06, 20:58:19