Kinematics for the ESAT

Scalar and Vector Quantities

A scalar quantity is defined as one that possesses magnitude (size) only, without direction. Examples include mass, time, energy, temperature, power, density, pressure, speed, and distance. It is important to note that a scalar can be negative. For example, a temperature below 0 degrees Celsius is negative, as is potential energy measured below an arbitrary reference point. However, some scalars like mass can never be negative, though their change can be.

A vector quantity possesses both magnitude and direction. Examples include force, velocity, acceleration, displacement, and momentum. Vectors are often defined relative to a specific direction. For instance, if upwards is defined as positive, a force acting downwards is negative. When a vector changes, that change is itself a vector with a direction. If an object moves at $5\text{ ms}^{-1}$ to the left and increases to $8\text{ ms}^{-1}$ to the left, the change is $3\text{ ms}^{-1}$ to the left.

Worked Example: Identifying Vectors

Consider this narrative: "It was -5 degrees Celsius. It took 10 minutes to defrost the car. I was 25 minutes late getting to work, which is 750m away in a direct line. I never exceeded 20 mph, though walking the 3 mile shorter footpath at 2 mph might have been faster." Which quantity is the magnitude of a vector?

The temperature and times are scalars. The figures 20 mph and 2 mph are speeds (scalars). The 3 mile figure is a distance (scalar). The 750m figure, representing a straight-line measurement from home to work, is displacement. Displacement is a vector, so 750m is the answer.

Worked Example: Displacement Directions

A person walks 5.0m north, 6.0m east, 7.0m south, and 8.0m west. What is the final displacement north of the starting point?

We ignore east and west components as they do not affect the north-south displacement. The person moves +5.0m (north) and -7.0m (south). The net displacement is $5.0 - 7.0 = -2.0\text{m}$.

Distance, Displacement, Speed, and Velocity

Distance is a scalar, while displacement is a vector. Displacement is distance in a specific direction. Similarly, speed is the scalar magnitude of velocity, which is speed in a given direction.

Circular Motion Example

Consider a car on a circular track with a radius of 100m and a circumference of 630m.

At the halfway point H, the car has travelled a distance of 315m, but its displacement from the start S is +200m. Upon returning to S, the distance is 630m, but the displacement is zero. On the second lap at point H, the distance is 945m, but the displacement remains +200m.

Suppose the car moves at a constant speed of $20\text{ ms}^{-1}$. At S, its velocity is $+20\text{ ms}^{-1}$ to the right. At H, its velocity is $-20\text{ ms}^{-1}$ to the right (meaning $+20\text{ ms}^{-1}$ to the left).

Worked Example: The Lift

A lift starts at floor 2, moves to floor 8, then 6, then 9, and finally floor 1. Each floor is 5.0m apart. What is the distance and displacement?

1. Up 6 floors (+30m).
2. Down 2 floors (10m).
3. Up 3 floors (+15m).
4. Down 8 floors (40m).

Total distance = $30 + 10 + 15 + 40 = 95\text{m}$. Net displacement = $30 - 10 + 15 - 40 = -5.0\text{m}$.

Worked Example: Bouncing Ball

A ball hits the ground at $15\text{ ms}^{-1}$ and rebounds at $10\text{ ms}^{-1}$. What is the change in speed and velocity (upwards is positive)?

Change in speed = $15 - 10 = 5\text{ ms}^{-1}$. Initial velocity = $-15\text{ ms}^{-1}$. Final velocity = $+10\text{ ms}^{-1}$. Change in velocity = $10 - (-15) = +25\text{ ms}^{-1}$.

Calculations of Speed and Velocity

The following formulae apply when speed or velocity is constant:

$\text{speed} = \frac{\text{distance}}{\text{time}}$

$\text{velocity} = \frac{\text{change in displacement}}{\text{time}}$

Using the circular track (630m circumference, 20s per lap): For a full lap: Speed = $630 / 20 = 31.5\text{ ms}^{-1}$. Velocity = $0 / 20 = 0\text{ ms}^{-1}$. For a half lap: Speed = $315 / 10 = 31.5\text{ ms}^{-1}$. Velocity = $200 / 10 = 20\text{ ms}^{-1}$ in the $S \to H$ direction.

Worked Example: Train Velocity

A train travels east at $80\text{ ms}^{-1}$ for 40 minutes from X to Y. It then travels west to Z (144km from X) in 16 minutes. What is the velocity from Y to Z?

Distance $XY = 80 \times 40 \times 60 = 192,000\text{m} = 192\text{km}$. Distance $XZ = 144\text{km}$, so distance $YZ = 192 - 144 = 48\text{km}$. Since Z is west of Y, displacement $YZ = -48\text{km}$. Velocity = $-48,000 / (16 \times 60) = -50\text{ ms}^{-1}$.

Acceleration

Acceleration is the rate of change of velocity:

$\text{acceleration} = \frac{\text{change in velocity}}{\text{time}}$

The unit is $\text{ms}^{-2}$. This formula applies only to constant acceleration. Note that an object can have acceleration even if its velocity is momentarily zero, or if it moves at a constant speed in a circle (because the direction changes).

Pendulum Motion Analysis

Consider a pendulum bob swinging between points 1 and 5 (right is positive):

Position	Velocity	Acceleration	Description
1	Zero	Positive	At rest, accelerating right
2	Positive	Positive	Moving right, speeding up
3	Positive	Zero	Max speed to the right
4	Positive	Negative	Moving right, slowing down
5	Zero	Negative	At rest, accelerating left

Worked Example: Acceleration of a Ball

A ball hits a bat at $25\text{ ms}^{-1}$ and rebounds at $20\text{ ms}^{-1}$ in the opposite direction. Impact time is 0.30s. Initial direction is positive.

Initial velocity = $+25\text{ ms}^{-1}$. Final velocity = $-20\text{ ms}^{-1}$. Change in velocity = $-20 - 25 = -45\text{ ms}^{-1}$. Acceleration = $-45 / 0.30 = -150\text{ ms}^{-2}$.

Interpreting Motion Graphs

1. Distance-time graphs: Always positive and always increasing (or flat). The gradient is the speed.
2. Displacement-time graphs: Can be positive or negative. The gradient is the velocity.
3. Speed-time graphs: Magnitude only. The area is the distance.
4. Velocity-time graphs: Shows direction. The gradient is acceleration, and the area is the change in displacement.

Displacement-Time Features

• Zero gradient: stationary.
• Positive/Negative gradient: moving in the positive/negative direction.
• Increasing/Decreasing gradient: increasing/decreasing velocity.

Velocity-Time Features

• Zero value: stationary.
• Positive/Negative value: moving in the positive/negative direction.
• Positive/Negative gradient: accelerating in the positive/negative direction.

Worked Example: Graph Interpretation

In this displacement-time graph (downwards is positive), the ball has an increasing downward speed at points Q, U, and Y (steeper positive gradient). It is momentarily at rest at P, T, and X.

In this velocity-time graph (downwards is positive), the ball is at the top of a bounce when velocity is zero at T and W.

Calculations using Gradients and Areas

Worked Example: Gradient

To find the velocity at $t = 35\text{s}$, calculate the gradient between 30s and 40s: Change in displacement = $0.0 - 5.0 = -5.0\text{ cm}$. Time = 10s. Velocity = $-5.0 / 10 = -0.50\text{ cm s}^{-1}$.

Worked Example: Area under a Velocity-Time Graph

For a ball hitting a bat, the net displacement is the sum of the areas:

Area 1 (Positive) = $0.5 \times (58 - 50) \times 10^{-3} \times 40 = +0.16\text{m}$.

Area 2 (Negative) = $0.5 \times (63 - 58) \times 10^{-3} \times (-25) = -0.0625\text{m}$.

Net displacement = $0.16 - 0.0625 = +0.0975\text{m} = +9.75\text{ cm}$.

Average Speed, Velocity, and Acceleration

• $\text{Average speed} = \frac{\text{total distance}}{\text{total time}}$
• $\text{Average velocity} = \frac{\text{net displacement}}{\text{total time}}$
• $\text{Average acceleration} = \frac{\text{net change in velocity}}{\text{total time}}$

For the graph above, distance in first 30s is 30m ($1.0\text{ ms}^{-1}$). Distance in next 20s is 70m ($3.5\text{ ms}^{-1}$). Average speed = $100 / 50 = 2.0\text{ ms}^{-1}$.

From the complex graph above: Total displacement = $25 + 100 + 25 - 16 - 24 = 110\text{m}$. Average velocity = $110 / 60 = 1.83\text{ ms}^{-1}$. Total distance = $25 + 100 + 25 + 16 + 24 = 190\text{m}$. Average speed = $190 / 60 = 3.17\text{ ms}^{-1}$. Average acceleration = $0$ (since $v_{\text{initial}}$ and $v_{\text{final}}$ are both $0$).

Equations of Uniform Motion

When acceleration is constant, we use the equations of motion:

1. $v = u + at$
2. $s = \frac{1}{2}(u + v)t$
3. $v^2 - u^2 = 2as$

These are derived from velocity-time graphs. The gradient is $(v-u)/t$, giving $v = u + at$. The area under the graph is the displacement $s = \frac{1}{2}(u+v)t$. Combining these leads to $v^2 - u^2 = 2as$.

Worked Example: Stone thrown from a Cliff

A stone is thrown up at $5.0\text{ ms}^{-1}$ from a 30m cliff. Acceleration is $10\text{ ms}^{-2}$ downwards. At what speed does it hit the sea? (Downwards is positive).

$u = -5.0\text{ ms}^{-1}$, $a = +10\text{ ms}^{-2}$, $s = +30\text{m}$. $v^2 - (-5.0)^2 = 2 \times 10 \times 30$ $v^2 - 25 = 600 \to v^2 = 625$ $v = 25\text{ ms}^{-1}$.

Worked Example: Car Braking

A car accelerates from $20\text{ ms}^{-1}$ to $30\text{ ms}^{-1}$ at $5.0\text{ ms}^{-2}$. Find the distance. $2 \times 5.0 \times s = 30^2 - 20^2 = 900 - 400 = 500$. $s = 500 / 10 = 50\text{m}$.