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# Velocity and Acceleration

## The speed and direction of a moving object and how that rate changes over time.

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Practice Velocity and Acceleration
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Velocity and Acceleration - May 03

Students will learn the meaning of acceleration, how it is different than velocity and how to calculate average acceleration.

### Key Equations

$v =$ velocity (m/s)

$v_i =$ initial velocity

$v_f =$ final velocity

$\Delta v =$ change in velocity $= v_f - v_i$

$v_{avg} = \frac{\Delta x}{\Delta t}$

$a =$ acceleration $(m/s^2)$

$a_{avg} = \frac{\Delta v}{\Delta t}$

Guidance
• Acceleration is the rate of change of velocity. So in other words, acceleration tells you how quickly the velocity is increasing or decreasing. An acceleration of $5 \ m/s^2$ indicates that the velocity is increasing by $5 m/s$ in the positive direction every second.
• Gravity near the Earth pulls an object downwards toward the surface of the Earth with an acceleration of $9.8 \ m/s^2 ( \approx 10 \ m/s^2)$ . In the absence of air resistance, all objects will fall with the same acceleration. The letter $g$ is used as the symbol for the acceleration of gravity.
• When talking about an object's acceleration, whether it is due to gravity or not, the acceleration of gravity is sometimes used as a unit of measurement where $1g=9.8m/s^2$ . So an object accelerating at 2g's is accelerating at $2*9.8m/s^2$ or $19.6m/s^2$
• Deceleration is the term used when an object’s speed (i.e. magnitude of its velocity) is decreasing due to acceleration in the opposite direction of its velocity.

#### Example 1

A Top Fuel dragster can accelerate from 0 to 100 mph (160 km/hr) in 0.8 seconds. What is the average acceleration in $m/s^2$ ?

Question: $a_{avg} = ? \ [m/s^2]$

Given: $v_i = 0 \ m/s$

${\;} \qquad \ \ v_f = 160 \ km/hr$

${\;} \qquad \ \quad t = 0.8 \ s$

Equation: $a_{avg} = \frac{\Delta v }{t}$

Plug n’ Chug: Step 1: Convert km/hr to m/s

$v_f = \left( 160 \frac{km}{hr} \right ) \left( \frac{1,000 \ m}{1 \ km} \right ) \left ( \frac{1 \ hr}{3,600 \ s} \right ) = 44.4 \ m/s$

Step 2: Solve for average acceleration:

$a_{avg} = \frac{\Delta v}{t} = \frac{v_f - v_i}{t} = \frac{44.4 \ m/s - 0 \ m/s}{0.8 \ s} = 56 \ m/s^2$

Answer: $\boxed {\mathbf{56 \ m/s^2}}$ Note that this is over $5 \frac{1}{2}$ g’s!

### Time for Practice

1. Ms. Reitman’s scooter starts from rest and accelerates at $2.0 m/s^2$ .
1. Where will the scooter be relative to its starting point after 7.0 seconds?
2. What is the scooter's velocity after 1s? after 2s? after 7s?
2. A horse is galloping forward with an acceleration of $3 \;\mathrm{m/s}^2$ . Which of the following statements is not necessarily true? You may choose more than one.
1. The horse is increasing its speed by 3 m/s every second, from 0 m/s to 3 m/s to 6 m/s to 9 m/s.
2. The speed of the horse will triple every second, from 0 m/s to 3 m/s to 9 m/s to 27 m/s.
3. Starting from rest, the horse will cover 3 m of ground in the first second.
4. Starting from rest, the horse will cover 1.5 m of ground in the first second.
3. Below are images from a race between Ashaan (above) and Zyan (below), two daring racecar drivers. High speed cameras took four pictures in rapid succession. The first picture shows the positions of the cars at $t = 0.0$ . Each car image to the right represents times 0.1, 0.2, and 0.3 seconds later.
1. Who is ahead at $t = 0.2 \;\mathrm{s}$ ? Explain.
2. Who is accelerating? Explain.
3. Who is going fastest at $t = 0.3 \;\mathrm{s}$ ? Explain.
4. Which car has a constant velocity throughout? Explain.
5. Graph $x$ vs. $t$ and $v$ vs. $t$ . Put both cars on same graph; label which line is which car.
6. Which car is going faster at $t = 0.2 \;\mathrm{s}$ (Hint: Assume they travel the same distance between 0.1 and 0.2 seconds)?

1. a. 49 m b. 2 m/s, 4 m/s, 14 m/s

2. discuss in class

3. See Video above