# Acceleration

11SCI - Mechanics

2020

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## What is Acceleration?

Acceleration is how quickly the velocity changes.

e.g. A supercar will accelerate to 50km/hr faster than a cyclist. That is to say, the supercar has a greater acceleration.

## Calculating Acceleration

\begin{aligned} acceleration &= \frac{\text{change in speed}}{\text{change in time}} \\ a &= \frac{\Delta v}{\Delta t} \\ \end{aligned}
• Velocity ($$v$$) has units meters per second ($$ms^{-1}$$)
• Time ($$t$$) has units seconds ($$s$$)
• Acceleration ($$a$$) has units meters per second per second ($$m/s^{2}$$ or $$ms^{-2}$$)

## Rearranging Equations

\begin{aligned} a &= \frac{\Delta v}{\Delta t} && \text{v is divided by t} \\ a \times \Delta t &= \Delta v && \text{Undo the divide by multiplying} \\ \Delta t &= \frac{\Delta v}{a} && \text{Undo the multiplication by dividing} \end{aligned} \begin{aligned} v &= \frac{\Delta d}{\Delta t} && \text{d is divided by t} \\ v \times \Delta t &= \Delta d && \text{Undo the divide by multiplying} \\ \Delta t &= \frac{\Delta d}{v} && \text{Undo the multiplication by dividing} \end{aligned}

## Question 1

• A bee starts at rest and takes off from a flower and reaches a velocity of $$0.75ms^{-1}$$ in $$0.5s$$. Calculate it’s acceleration.

## Question 1: Answer

1. Knowns: $$\text{time (t)} = 0.5s$$, $$\Delta v = v_{f} - v_{i} = 0.75 - 0 = 0.75ms^{-1}$$
2. Unknowns: $$\text{acceleration (a)}$$
3. Formula: $$a = \frac{\Delta v}{\Delta t}$$
4. Substitute: $$a = \frac{0.75}{0.5}$$
5. Solve: $$v = 1.5\frac{m}{s^{2}} = 1.5ms^{-2}$$

## Question 2

A Bugatti Veyron accelerates from rest at $$27.77ms^{-2}$$ for $$2.6s$$. How fast is it travelling at after 2.6 seconds?

## Question 2: Answer

1. Knowns: $$\text{time (t)} = 2.6s$$, $$\text{acceleration (a)} = 27.77ms^{-2}$$
2. Unknowns: $$\text{final velocity } v_{f}$$
3. Formula: $$a = \frac{\Delta v}{\Delta t} = \frac{v_{f} - v_{i}}{\Delta t}$$
4. Substitute: $$27.77 = \frac{v_{f} - 0}{2.6}$$
5. Solve: $$27.77 \times 2.6 = v_{f} = 72.2ms^{-1}$$

## Question 3

A skydiver at rest jumps out of a plane. They accelerate at $$9.8ms^{-2}$$ until they reach a terminal velocity of $$54ms^{-2}$$. How long does it take them to reach this speed?

## Question 3: Answer

1. Knowns: $$\text{acceleration (a)} = 9.8ms^{-2}$$, $$\text{final velocity } (v_{f}) = 45ms^{-1}$$
2. Unknowns: $$\text{time (t)}$$
3. Formula: $$a = \frac{\Delta v}{\Delta t} = \frac{v_{f} - v_{i}}{\Delta t}$$
\begin{aligned} 9.8 &= \frac{54 - 0}{t} \\ 9.8 \times t &= 54 \\ t &= \frac{54}{9.8} = 5.51s \end{aligned}

## Question 4

A runner is approaching the finish line, moving at $$5.55ms^{-1}$$ but needs to sprint to pass the person just in front of them to get 1st place. They accelerate for $$3s$$ to reach $$6.3ms^{-1}$$. What is their acceleration?

## Question 4: Answer

1. Knowns: $$v_{i} = 5.55ms^{-1}$$, $$v_{f} = 6.3ms^{-1}$$, $$\Delta t = 3s$$
2. Unknowns: $$\text{acceleration (a)}$$
3. Formula: $$a = \frac{\Delta v}{\Delta t} = \frac{v_{f} - v_{i}}{\Delta t}$$
\begin{aligned} a &= \frac{6.3 - 5.55}{3} \\ a &= \frac{0.75}{3} = 0.25ms^{-2} \end{aligned}

## Mahi Kāinga

• Due Monday, August 24th: Mahi Kāinga Booklet Q2, Q3, Q1
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