Speed, Velocity and Acceleration

Distance and Displacement

Distance is defined as the total length taken between two points

- It is a scalar. meaning it is a quantity having magnitude only.

Displacement is defined as the change of position of a point in a particular direction

- It is a vector. meaning it is quantity having both magnitude and direction

SI units of both distance and displacement are metre [m].

Example 1

A car moves 5km to the East and 3km to the North. What is the distance and the displacement of the car?

 Distance and Displacement

Solutions

- Distance of the car is 8km (= 5km +3km).

- Displacement of the car is 5km East and 3km North.

Example 2

The circumference of a roundabout is 10m and the car turns it once. What is the distance and the displacement of the car?

 Distance and Displacement example 1

Solutions

- Distance of the car is 10m

- Displacement of the car is 0m because it came back to the starting position

Example 3

You walk forward 15m and backward 5m. What is your distance and your displacement?

 Distance and Displacement example 2

Solutions

Your distance is 20m (= 15m + 5m).

Your displacement is 10m (= 15m - 5m) forward.

Speed

Speed is defined as the rate of change of distance traveled with time. It is a scalar.

The unit of speed is metre per second [m/s].

Formula for Speed

Speed = Distance traveled Time taken

Average Speed = Total distance traveled Total time taken

Example

A car travels a distance of 540km from Lusaka to Katete in 10 hours. Find the average speed in km/hr and m/s.

Solution

Average Speed = Total distance traveled Total time taken

given that:

Total distance traveled = 540km

Total time taken = 10hrs

Average Speed = 540 10

Answer: Average Speed = 54 Km/hr

or Average Speed = 540000m 36000s

Answer Average Speed = 15 m/s

Velocity

Velocity is defined as the rate of change of displacement with time. It is a vector.

The unit of velocity is metre per second [m/s]. (It is the same unit as speed.)

Example

What are their speeds and their velocities?

Velocity  example 1

Solution

They have the same speeds of 10m/s but they have different velocities.

Car1 has the velocity of 10m/s East.

Car2 has the velocity of 10m/s North.

Acceleration

Acceleration is defined as the rate of change of velocity with time. It is a vector.

The unit of acceleration is metre per second squared [m/s2].

Formula for Acceleration

a = V - U t

a: Acceleration [m/s2]

v: final velocity [m/s]

u: initial velocity [m/s]

t: time taken [s]

Example 1

A car starting from rest increases its velocity uniformly to 15m/s in 5s. What is its acceleration?

Solution

Given that

V = 15 m/s

U = 0 m/s

t = 5s

a = V - U t

a = 15 - 0 5

a = 15 5

Answer: a = 3 m/s 2

Example 2

If a car slows down from 72km/hr and stops in 10s, calculate the acceleration

Solution

Convert 72km/hr to m/s

72 km 1 hr

72000m 3600

20 1

20m/s

Therefore the data is:

V = 0m/s

U = 20m/s

t = 10s

a = V - U t

a = 0 - 20 10

a = - 20 10

Answer: a = - 2 m/s 2

When the velocity reduces, the acceleration becomes a negative number. The acceleration is called the retardation or deceleration.

Uniformly accelerated liner motion

If a body moves with a uniform acceleration (the acceleration is constant), three important equations are given below.

Formulas for uniform acceleration

1. V = u + at

2. x = ut + 1 2 at 2

3. V 2 = u 2 + 2ax

Note that:

a: Acceleration [m/s2]

v: final velocity [m/s]

u: initial velocity [m/s]

t: time taken [s]

x: distance covered [m]

Example 1

A car traveling at 10m/s accelerates at 2m/s2 for 3s. What is its final velocity?

Solution

Formula

V = u + at

Given that

U = 10m/s

t = 3s

a = 2m/s 2

V = ?

V = 10 + 2(3)

V = 10 + 6

Answer: V = 16m/s

Example 2

A motorcycle starting from rest acquires a velocity of 72km/hr in 5s.

(a) What is its acceleration?

(b) How far does it travel during this time?

Solutions

(a)

Given that

V = 72km/hr = 20m/s

U = 0m/s

t = 5s

a = ?

a = V - U t

a = 20 - 0 5

a = 20 5

Answer: a = 4m/s 2

(b)

Given that

u = 0m/s

t = 5s

a = 4m/s2

x = ?

x = ut + 1 2 at 2

x = 0(5) + 1 2 4(5) 2

x = 0 + 1 2 4 (25)

x = 1 2 100

x = 50m

Note that: You can't use the formula: D=S×T. Because there is no acceleration.

Example 3

The velocity of an object is uniformly reduced from 50m/s to 30m/s. If the deceleration is –4m/s2, how much is the distance of the body decelerating?

Solution

Given that

u = 50m/s

v = 30m/s

a = -4m/s2

x = ?

V 2 = u 2 + 2ax

30 2 = 50 2 + 2(-4)x

900 = 2500 - 8x

900 - 2500 = - 8x

- 1600 = - 8x

- 8x -8 = - 1600 -8

x = 200m

Acceleration due to gravity

All objects accelerate uniformly towards the earth if air resistance is ignored. It is called acceleration due to gravity. It is represented by the symbol ‘g’.

g = 9.8m/s2 ≈ 10m/s2

Points to note

1. If a stone is dropped from the top of a tall building, it accelerates uniformly downwards. If you release a stone without applying force, it starts from rest. It is called free fall.

Free fall

u = 0m/s

a = g = 10m/s2

2. If you throw up a stone, the stone decelerates to the top. Then it stops momentarily at the top. And then it starts falling freely

Throwing up

v = 0m/s

a = -g = -10m/s2

Example 1

A body falls freely from rest. Air resistance is ignored. (g = 10m/s2)

(a) What is its velocity after 1s?

(b) How far does it reach in 1s?

Solutions

(a)

Given that

u = 0m/s

a = g = 10m/s2

t = 1s

v = ?

V = u + at

V = 0 + 10(1)

Answer: V = 10m/s

(b)

Given that

u = 0m/s

a = g = 10m/s2

t = 1s

x = ?

x = ut + 1 2 at 2

x = 0(1) + 1 2 10(1) 2

x = 0 + 1 2 10(1)

x = 1 2 10

Answer: x = 5m

Speed (velocity) – time graph

Speed (velocity) – time graphs tell stories about the movement of an object.

- The gradient of the speed – time graph is equal to the acceleration of the object.

- The area under the speed – time graph represents the distance traveled by the object.

The diagrams below show the speed – time graphs for different kinds of motion.

Speed (velocity) – time graph example 1

Example

A car moving from rest acquires a velocity of 20m/s with uniform acceleration in 4s. It moves with this velocity for 6s and again accelerates uniformly to 30m/s in 5s. It travels for 3s at this velocity and then comes to rest with uniform deceleration in 12s.

(a) Draw a speed – time graph

(b) Calculate the total distance covered.

(c) Calculate the average speed.

Solutions

(a)

Speed (velocity) – time graph example 2

(b)

To find the total distance covered, calculate the area under the speed – time graph.

Area A = triangle = 1 2 bh 1 2 4 × 20 = 40

Area B = rectangle = lb = 6 × 20 = 120

Area C = trapezium = 1 2 (a + b) h = 1 2 (20 + 30) 5 = 125

Area D = rectangle = lb = 3 × 30 = 90

Area E = triangle = 1 2 bh 1 2 12 × 30 = 180

Total distance covered = Total area = 40 + 120 + 125 + 90 + 180 = 555m

(c)

Average speed = Total distance covered Total time taken

Average speed = 555 30

Answer: Average speed = 18.5m/s