**Units Used This Topic:**

**area **metre-squared (m^{2})

**distance** metre (m)

**acceleration** metre/second-squared (m/s^{2})

**speed ** metre/second (m/s)

**velocity** metre/second (m/s)

**time ** second (s)

**Speed** does not involve direction. Speed is a** scalar** quantity.

The speed of a moving object is rarely constant. When people walk, run or travel in a car their speed is constantly changing.

The speed at which a person can walk, run or cycle depends on many factors including: age, terrain, fitness and distance travelled.

Typical values may be taken as:

- walking ̴1.5 m/s
- running ̴3 m/s
- cycling ̴6 m/s.

A typical value for the speed of sound in air is 330 m/s.

For an object moving at constant speed the distance travelled in a specific time can be calculated using the equation:** **

**distance travelled = speed × time**

**Question: **Calculate the speed of an object that moves 4m in 3s

**distance travelled = speed × time**

4 = speed x 3

speed = 4 / 3

The speed of the object is 1.3 m/s

**Question: **The thinking distance is the distance travelled during the driver’s reaction time.

A car was travelling at 13 m/s

The driver’s reaction time was 0.6 s

**distance travelled = speed × time**

distance travelled = 13 x 0.6

The thinking distance is **7.8 m**

The **velocity** of an object is its speed in a given direction. Velocity is a** vector** quantity.

** Displacement** is distance with a direction. Displacement is a vector.

Acceleration is a change in speed and/or direction.

learn this definition

Motion in a circle involves constant speed but changing velocity because the direction is always changing

If an object moves along a straight line, the distance travelled can be represented by a** distance–time graph**.

## Distance-Time Graphs

The speed of an object can be calculated from the **gradient** of its distance–time graph.

If an object is accelerating, its speed at any particular time can be determined by drawing a** tangent **and measuring the gradient of the distance–time graph at that time.

**Question:** The graph below shows a distance-time graph for 50 seconds of a bicycle ride.

The gradient of the distance-time graph gives the speed of the bicycle.

Determine the speed of the bicycle.

The speed is the gradient of the line

gradient = rise/run

gradient = 250 – 0 / 50 – 0

The speed = 5 m/s

**Question: **The graph shows a distance-time graph for an athlete in a race.

Describe how you can use the graph to determine the velocity of the athlete 20 minutes after the start of the race.

draw a tangent

at 20 minutes

measure the gradient of the tangent

## Average Acceleration (unit: m/s^{2})

The **average acceleration** of an object can be calculated using the equation:

**acceleration = change in velocity / time taken**

or

**acceleration = (final velocity – initial velocity) / time taken**

An object that slows down is** decelerating**.

**Question:** In a race, the athlete had a constant acceleration during the first 3.2 seconds. His velocity increased from 0 m/s to 11.6 m/s

Calculate the acceleration of the athlete.

**acceleration = (final velocity – initial velocity) / time taken**

acceleration = (11.6 – 0)/3.2

The acceleration is** 3.625m/s ^{2}**

## Uniform Acceleration over a Distance (unit: m/s^{2})

The following equation applies to uniform acceleration:

**final velocity² − initial velocity² = 2 × acceleration × distance**

**Question:** A lorry travels 84 m with a constant acceleration of 2.0 m/s^{2} to reach a velocity of 19 m/s

Calculate the initial velocity of the lorry.

**final velocity² − initial velocity² = 2 × acceleration × distance**

19^{2} – initial velocity^{2} = 2 x 2 x 84

361- initial velocity^{2} = 336

361 – 336 = initial velocity^{2}

25 = initial velocity^{2}

The initial velocity is **5 m/s**

## Velocity-Time Graphs

The** acceleration** of an object can be calculated from the **gradient** of a velocity–time graph.

The **distance travelled** by an object (or displacement of an object) can be calculated from the **area under** a velocity–time graph.

The distance travelled by an object (or displacement of an object) can be calculated from the area under a velocity–time graph.

If the graph is not a straight line, you can count the squares to find the area

**Question:** The graph shows how the velocity of a car changes with time.

Which feature of the graph represents the negative acceleration of the car?

The gradient of the line

**Question**: The graphs show how the velocity of two cars, **A** and **B**, change from the moment the car drivers see an obstacle blocking the road.

Time in secondsTime in seconds

One of the car drivers has been drinking alcohol. The other driver is wide awake and alert.

Use the graphs to calculate how much further car **B** travels before stopping compared to car **A**.

Distance travelled by car A:

= (15 x 0.6) + (0.5 x 15 x 2.6)

distance travelled by A = 28.5 m

Distance travelled by car B:

= (15 x 1.4) + (0.5 x 15 x 2.6)

distance travelled by B = 40.5.

Extra distance travelled by B = 40.5 – 28.5 = **12m**

## Falling Under Gravity

Near the Earth’s surface any object falling freely under gravity has an acceleration of about **9.8 m/s ^{2}**

An object falling through a fluid initially accelerates due to the force of gravity.

As the falling object increases speed, the air resistance on it increases

When the air resistance = the weight, the forces acting on the object are balanced: there is no resultant force.

Eventually the **resultant force will be zero** and the object will move at its **terminal velocity**.

**Question: **The graph shows how the vertical velocity of a parachutist changes from the moment the parachutist jumps from the aircraft until landing on the ground.

Using the idea of forces, explain why the parachutist reaches a terminal velocity and why opening the parachute reduces the terminal velocity. (6 marks)

**first terminal velocity**

- on leaving the plane the only force acting is weight (downwards)
- as parachutist falls air resistance acts (upwards)accept drag / friction for air resistance
- weight greater than air resistance
- as velocity / speed increases so does air resistance
- terminal velocity reached when air resistance = weight

**to explain second lower terminal velocity**

- opening parachute increases surface area
- which air resistance
- air resistance is greater than weight
- resultant force acts upwards / opposite direction to motion
- parachutist decelerates / slows down
- the lower velocity means a reduced air resistanceair resistance and weight become equal but at a lower (terminal) velocity