Acceleration

11SCI - Mechanics

Finn LeSueur

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
  • Tutorials: Tuesday Lunchtime in A5
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  • Do the work in your exercise book.