U3AOS1 Topic 1: Kinematics

Scalar vs Vector

Scalar quantities only have a magnitude.
Vector quantities have a magnitude AND a direction

Displacement vs Distance

Displacement is how far you are from where you started
Distance is how far you travel to get there

Suppose you want to travel from the MCG to Mt Buller. The car route is the distance you would need to travel to get there.
All the car routes are 230km+ but the straight black line is the displacement which is only 150km.

Average Speed vs Velocity

Speed is a scalar quantity related to distance and time
Velocity is a vector quantity related to displacement and time

$ d$ = distance, $ t$=time, $s $ = displacement
\[ \displaystyle \Large speed_{\text{avg}}= \frac{d}{t}\]
\[ \displaystyle \Large \text{velocity}= v= \frac{\Delta s}{\Delta t}\]
$speed_{average} = \frac{d}{t}$

IMPORTANT INFO
A common mistake in kinematics is forgetting that
IF VELOCITY IS CONSTANT THEN ACCELERATION IS 0

Justification:
- Acceleration is the rate of change of velocity
- If the velocity is not changing (constant)
- Then will be 0

Plugging this into the formula gives
\[ \displaystyle \Large a= \frac{\Delta v}{\Delta t}\]
\[ \displaystyle \Large a=\frac{0}{\Delta t}\]
\[ \displaystyle \Large a=0\]

$a = \frac{Δ v}{Δ t}$

SUVAT EQUATIONS
\[ \displaystyle \Large v=u+at\]
\[ \displaystyle \Large v^2=u^2+2as\]
\[ \displaystyle \Large s= ut + \frac{1}{2}at^2\]
\[ \displaystyle \Large s=vt - \frac{1}{2}at^2\]
\[ \displaystyle \Large s=\frac{1}{2}(u+v)t\]

$v = u+at$

The table below simplifies SUVAT and is really good to have in your cheat sheet to easily find which formula to use.

SUVAT CALCULATOR

Use the SUVAT calculator below to check your answers (enter 3 or more variables)

CONVERTING KM/H TO M/S

To go from $km/h \rightarrow m/s$ DIVIDE by $3.6$

To go from $m/s \rightarrow km/h$ MULTIPLY by $3.6$
$3 .$

Example 1

Hank drives 120 km in 2 hours at a constant speed.

1. What is Hank’s average speed?

2. What is Hank’s acceleration?

Solution to Question 1.

Identify what is given:

$120 k m$ is the distance and $2 h o u r s$ is the time.

The question asks for the average speed which has the formula of

$a v g s p e e d = \frac{d}{t}$

Substituting $� = 120$ and $� = 2$ gives

$a v g s p e e d = \frac{120 k m}{2 h o u r s} = \frac{60 k m}{1 h o u r} = 60 k m / h$

Solution to Question 2.

As Hank travels at a constant speed the acceleration will be 0.

This is because acceleration is the rate of change of velocity and as the speed is not changing the acceleration will remain 0.

Exercise &&1&& (&&5&& Questions)

A car travels at a constant speed of 60 km/h for 2 hours. Calculate the total distance covered by the car.

Submit

A person walks 500 meters east, then turns and walks 300 meters north. Determine the total displacement from the starting point.

Submit

A boat travels upstream for 2 hours at a speed of 5 m/s and then downstream for 3 hours at a speed of 8 m/s. Calculate the total distance covered by the boat.

Submit

A cyclist travels at a speed of 15 km/h for 1.5 hours. Calculate the distance traveled by the cyclist.

Submit

A cat runs in a circle at a speed of 8 m/s for 10 seconds and returns to its starting position. Calculate the displacement of the cat.

Submit

U3AOS1 Topic 1: Kinematics

Hank drives 120 km in 2 hours at a constant speed.

1. What is Hank’s average speed?

2. What is Hank’s acceleration?

Solution to Question 1.

Identify what is given:

$120 k m$ is the distance and $2 h o u r s$ is the time.

The question asks for the average speed which has the formula of

$a v g s p e e d = \frac{d}{t}$

Substituting $� = 120$ and $� = 2$ gives

$a v g s p e e d = \frac{120 k m}{2 h o u r s} = \frac{60 k m}{1 h o u r} = 60 k m / h$

Solution to Question 2.

As Hank travels at a constant speed the acceleration will be 0.

This is because acceleration is the rate of change of velocity and as the speed is not changing the acceleration will remain 0.

A car travels at a constant speed of 60 km/h for 2 hours. Calculate the total distance covered by the car.

A person walks 500 meters east, then turns and walks 300 meters north. Determine the total displacement from the starting point.

A boat travels upstream for 2 hours at a speed of 5 m/s and then downstream for 3 hours at a speed of 8 m/s. Calculate the total distance covered by the boat.

A cyclist travels at a speed of 15 km/h for 1.5 hours. Calculate the distance traveled by the cyclist.

A cat runs in a circle at a speed of 8 m/s for 10 seconds and returns to its starting position. Calculate the displacement of the cat.

Confident?

Select quiz difficulty

U3AOS1 Topic 1: Kinematics

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Hank drives 120 km in 2 hours at a constant speed.

1. What is Hank’s average speed?

2. What is Hank’s acceleration?

Solution to Question 1.

Identify what is given:

120km120km120km is the distance and 2hours2 hours2 hours is the time.

The question asks for the average speed which has the formula of

avgspeed=dtavg speed = \frac{d}{t} avg speed=td​

Substituting �=120d=120 and �=2t=2 gives

avgspeed=120km2hours=60km1hour=60km/havg speed = \frac{120km}{2 hours} =\frac{60km}{1hour}=60km/havg speed=2 hours120km​=1 hour60km​=60km/h

Solution to Question 2.

As Hank travels at a constant speed the acceleration will be 0.

This is because acceleration is the rate of change of velocity and as the speed is not changing the acceleration will remain 0.

A car travels at a constant speed of 60 km/h for 2 hours. Calculate the total distance covered by the car.

A person walks 500 meters east, then turns and walks 300 meters north. Determine the total displacement from the starting point.

A boat travels upstream for 2 hours at a speed of 5 m/s and then downstream for 3 hours at a speed of 8 m/s. Calculate the total distance covered by the boat.

A cyclist travels at a speed of 15 km/h for 1.5 hours. Calculate the distance traveled by the cyclist.

A cat runs in a circle at a speed of 8 m/s for 10 seconds and returns to its starting position. Calculate the displacement of the cat.

Confident?

Select quiz difficulty

$120 k m$ is the distance and $2 h o u r s$ is the time.

$a v g s p e e d = \frac{d}{t}$

Substituting $� = 120$ and $� = 2$ gives

$a v g s p e e d = \frac{120 k m}{2 h o u r s} = \frac{60 k m}{1 h o u r} = 60 k m / h$