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## Te Whāinga Ako

1. Review basic speed and acceleration calculations.

Write the date and te whāinga ako in your book

## Pātai Tahi: Who is the fastest?

• Andy can run $100m$ in $11.9$ seconds
• Bob can run $100m$ in $10.8$ seconds
• Chris can run $100m$ in $12.4$ seconds

### Whakatika

• Andy can run $100m$ in $11.9$ seconds
• Bob can run $100m$ in $10.8$ seconds
• Chris can run $100m$ in $12.4$ seconds

Bob because he ran $100m$ in the shortest time.

## Pātai Rua: Who is the fastest?

• Aaron can run $534m$ in $1 minute$
• Billy can run $510m$ in $1 minute$
• Cameron can run $452m$ in $1 minute$

### Whakatika

• Aaron can run $534m$ in $1 minute$
• Billy can run $510m$ in $1 minute$
• Cameron can run $452m$ in $1 minute$

Aaron because he ran the furthest in $1 minute$.

## Who is the fastest?

• Ash can run $0.3km$ in $45 seconds$
• Bailey can run $420m$ in $1 minute$
• Caleb can run $510m$ in $1.5 minutes$

## Average Speed

\begin{aligned} & v = \frac{d}{t} \newline & d = \text{total distance travelled} \newline & t = \text{time} \newline & v = \text{speed} \end{aligned}

Write this equation in your book and give the unit for each letter in the equation.

### What is the Unit?

• $ms^{-1}$
• It stands for meters per second
• E.g. the speed of sound is $343ms^{-1}$
• Sound travels $330m$ in one second

### Example / Tauria

Ash runs $315m$ in $45s$. Calculate his average speed in meters per second.

1. Knowns:
2. Unknowns:
3. Formula:
4. Substitute:
5. Solve:

#### Whakatika

Ash runs $315m$ in $45s$. Calculate his average speed in meters per second.

\begin{aligned} & d = 315m, t = 45s \newline & v = ? \newline & v = \frac{d}{t} \newline & v = \frac{315}{45} \newline & v = 7ms^{-1} \end{aligned}

### The Speed Of

• A skydiver (freefall) = $53ms^{-1}$
• A handgun bullet = $660ms^{-1}$
• A car on the road = $50km/hr$
• A flying airplane = $1100kmh^{-1}$
• Light = $300,000,000$

Pātai: In pairs, convert the speed of an airplane to meters per second.

#### Whakatika

\begin{aligned} v &= \frac{1100km}{hr} \newline &= \frac{1100km \times 1000}{60 \times 60} \newline &= \frac{1100000}{3600} = 305.56ms^{-1} \end{aligned}

### Pātai

A car is moving at a speed of $10ms^{-1}$. How far does the car travel in $12s$?

1. Knowns:
2. Unknowns:
3. Formula:
4. Substitute:
5. Solve:

#### Whakatika

A car is moving at a speed of $10ms^{-1}$. How far does the car travel in $12s$?

\begin{aligned} v &= 10ms^{-1}, t=12s \newline d &= ? \newline v &= \frac{d}{t} \newline 10 &= \frac{d}{12} \newline 10 \times 12 &= d = 120m \end{aligned}

### Pātai

A man is running at a speed of $4ms^{-1}$. How long does he take to run $100m$?

1. Knowns:
2. Unknowns:
3. Formula:
4. Substitute:
5. Solve:

#### Whakatika

A man is running at a speed of $4ms^{-1}$. How long does he take to run $100m$?

\begin{aligned} v &= 4ms^{-1}, d=100m \newline t &= ? \newline v &= \frac{d}{t} \newline 4 &= \frac{100}{t} \newline 4 \times t &= 100 \newline t &= \frac{100}{4} = 25s \end{aligned}

### Average vs Instantaneous Velocity

Velocity may refer to average velocity or instantaneous velocity.

The formula $v = \frac{d}{t}$ can only be used to calculate average velocity or when the velocity is constant.

## Ngā Whāinga Ako

1. Review basic acceleration calculations.
2. Cover a basic introduction vectors.

Write the date and ngā whāinga ako in your book

## Acceleration

The rate of change in speed

\begin{aligned} & a = \frac{\Delta v}{t} \newline & \Delta v = \text{ change in speed} \newline & t = \text{ time} \newline & a = \text{ acceleration} \end{aligned}

### What does $ms^{-2}$ mean?

meters per second squared OR meters per second per second

For example, $a=12ms^{-2}$ means that the velocity is increased by $12ms^{-1}$ every second.

### $\Delta$ = Delta

This is the difference between the initial and the final value.

\begin{aligned} & \Delta = final - initial \newline & \text{e.g. }\Delta v = v_{f} - v_{i} \end{aligned}

### Pātai

A man initially walking at $2.0ms^{-1}$ notices that his house is on fire so he speeds up to $11ms^{-1}$ in $1.3s$.

1. Calculate the change in speed
2. Calculate his acceleration

### Whakatika 1

\begin{aligned} & v_{f} = 11ms^{-1}, v_{i} = 2ms^{-1} && \text{Knowns}\newline & \Delta v = ? && \text{Unknowns}\newline & \Delta v = v_{f} - v_{i} && \text{Formula}\newline & \Delta v = 11 - 2 = 9ms^{-1} && \text{Sub and Solve} \end{aligned}

### Whakatika 2

\begin{aligned} & \Delta v = 9ms^{-1}, t = 1.3s && \text{Knowns} \newline & a = ? && \text{Unknowns} \newline & a = \frac{\Delta v }{t} && \text{Formula} \newline & a = \frac{9}{1.3} = 6.9ms^{-2} && \text{Sub and Solve} \end{aligned}

### Akoranga 2 Mahi Tuatahi

A cyclist who has been travelling at a steady speed of $4ms^{-1}$ starts to accelerate. If he accelerates at $2.5ms^{-2}$, how long will he take to reach a speed of $24ms^{-1}$?

K,U,F,S,S

#### Whakatika

\begin{aligned} & v_{i} = 4ms^{-1}, v_{f} = 24ms^{-1}, a = 2.5ms^{-2} && \text{Knowns} \newline & t = ? && \text{Unknowns} \newline & a = \frac{\Delta v}{t} && \text{Formula} \newline & t = \frac{\Delta v}{a} && \text{Rearrange by swapping a and t} \newline & t = \frac{v_{f}-v_{i}}{a} && \text{Expand Δv} \newline & t = \frac{24 - 4}{2.5} && \text{Substitute} \newline & t = 8s && \text{Solve} \end{aligned}

### More Pātai

1. A car initially moving at $12.7ms^{-1}$ accelerates at $1.3ms^{-2}$ for one minute. What is the car’s final speed?
2. A car decelerates at $1.8ms^{-2}$ for $9.4s$ to stop. What was the car’s initial speed?

Whakatika 1

\begin{aligned} & v_{i} = 12.7ms^{-1}, a = 1.3ms^{-2}, t = 60s && \text{Knowns} \newline & v_{f} = ? && \text{Unknowns} \newline & a = \frac{v_{f} - v_{i}}{t} && \text{Formula} \newline & a \times t = v_{f} - v_{i} && \text{Rearrange for final v} \newline & v_{f} = (a \times t) + v_{i} \newline & v_{f} = (1.3 \times 60) + 12.7 = 90.7ms^{-1} && \text{Sub and solve} \end{aligned}

Whakatika 2

\begin{aligned} & a = -1.8ms^{-2}, t = 9.4s, v_{f} = 0ms^{-1} && \text{Knowns} \newline & v_{i} = ? && \text{Unknowns} \newline & a = \frac{v_{f} - v_{i}}{t} && \text{Formula} \newline & a \times t = v_{f} - v_{i} && \text{Rearrange for initial v} \newline & v_{i} = v_{f} - (a \times t) \newline & v_{i} = 0 - (-1.8 \times 9.4) = 16.92ms^{-1} && \text{Sub and solve} \end{aligned}

Discuss with the person next to you, the relevance of the positive and negative signs.