In this tutorial, you will learn the following:

- Concepts of displacement and distance
- Distance-time, velocity-time and acceleration-time graphs
- How to convert a distance time graph to corresponding speed-time and acceleration-time graphs
- Average speed and average velocity
- Deriving v = u + at, s = ut + 1/2 at
^{2}and v^{2}= u^{2}+ 2as - Instantaneous velocity - interactive
- A good collection of worked examples

**Distance** is the amount of change in position of an object.

**Displacement** is the amount of change in position of an object in a certain direction.

Distance is a scalar and displacement is a vector. Displacement is equal to the shortest distance between two points in that direction.

There are four paths between points **A** and **B**. The shortest path is the displacement between the two.

The following animation help you distinguish between *displacement* and *distance.*

The rate of change of displacement - distance in a certain direction - is called the velocity.

Velocity = displacement /time

The **gradient** of a distance time graph is **speed.**

*Units: ms ^{-1}*

The rate of change of velocity - speed in a certain direction - is called the acceleration or deceleration.

Acceleration = velocity / time

The **gradient** of a velocity time graph is **acceleration or deceleration.**

*Units: ms ^{-2}*

The following animations show displacement / time graphs and their corresponding velocity / time graphs and acceleration / time graphs.

In the animations, please focus on the gradients of displacement-time graphs and velocity-time graphs, that will help understand the relationships.

**Average speed** is defined as the total distance travelled by an object divided by the time taken for it.

Average Speed = Total Distance / Total Time

**Average Velocity** is defined as the total displacement travelled by an object divided by the time taken for it.

Average Velocity = Total Displacement / Total Time

**E.g.1**

The distance between Hounslow and Kingston is 9km. A man walks from Hounslow to Kingston at 4km/h, takes a rest at the latter for an hour and comes back at 6km/h. Find the average speed and velocity of the man.

Time taken for Hounslow to Kingston = 9/4 = 2.25 hrs

Time taken for Kingston to Hounslow = 9/6 = 1.5 hrs

Total distance = 18km

Total time= 2.25 + 1 + 1.5 = 4.75

Average speed = 18/4.75

= 3.8km/h

Total displacement = 0

Total time = 4.74

Average velocity = 0/4.75 = 0.

**E.g.2**

A cyclist travels one half of the distance to his destination at 10km/h and the other half at 15km/h. Find the average speed.

d, t_{1}, 10km/h

d, t_{2}, 15km/h

t_{1} = d/10

t_{2} = d/15

Total time = d/10 + d/15 = d[1/10 + 1/15] = d[25/150] = d/6

Total distance = 2d

Average speed = 2d/(d/6) = 12km/h.

**E.g.3**

A cyclist travels one half of the time taken to his destination at 20km/h and the other half at 30km/h. Find the average speed.

t, d_{1}, 20km/h

t, d_{2}, 30km/h

d_{1} = 20t

d_{2} = 30t

Total time = 2t

Total distance = 20t + 30t = 50t

Average speed = 50t/2t = 25km/h.

**E.g.4**

A, B and C are three towns. The distance between A and B is 60km and the distance between B and C is 90km. A man drives from B to C at 30km/h and then back to A at 15km/h. Find the average speed and velocity of the car.

A

B

C

Time taken for from B to C = 90/30 = 3hrs

Time taken for from C to A = 150/15 = 10hrs

Total time = 13 hrs

Total distance = 240km

Average speed = 240/13

= 18.5 km/h

Total displacement = -60km

Average velocity = -60/13 = 4.6

= 4.6km/h, towards A.

From the graph, a = v-u/t

v = u + at 1

From the graph, the area, s = ut + 1/2 (v-u)t

s = ut + 1/2 at^{2} 2

From 1 => v^{2} = u^{2} + 2uat + a^{2}t^{2}

v^{2} = u^{2} + 2a(ut + 1/2 at^{2})

v^{2} = u^{2} + 2as 3

Please use the buttons to drop the ball and reset.

The following animation models a ball thrown upwards until it comes back down and hits the ground.

The velocity of the ball comes down, then becomes instantaneously zero and increases again. The corresponding dis;placement-time and acceleration-time graphs are shown on the same grid.

If the displacement-time graphs is a straight line, the gradient at any point gives the velocity. If it is not a straight line, then, its gradient changes from point to point and so does the velocity of the object in question. In situations like this, we find the **instantaneous** speed by drawing a tangent at the point in order to find the gradient. The gradient gives the velocity at the point - instantaneous velocity.

With the following applet, you can practise it interactively. Move the *slider* - the time - and see the instantaneous velocity of the point of your choice.

**E.g.**

This problem is based on the above applet. You may check your answers with the above.

The displacement of an object against time is given by the formula, s = 4 - (t-2)^{2}. Find the following:

- Find an expression for the instantaneous velocity at any time.
- Hence, find the instantaneous velocity, when t = 1s, t = 3s and t = 5s.
- Find the time, when the instantaneous velocity is zero.

- s = 4 - (t-2)
^{2}= 4 - [t^{2}- 4t + 4] = 4 - t^{2}+ 4t - 4 = 4t - t^{2}

v = ds/dt = 4 - 2t

v = 4 - 2t - t = 1 => v = 4 - 2 = 2m/s

t = 3 => v = 4 - 3 = -2m/s

t = 5 => v = 4 - 10 = -6m/s - v = 4 - 2t

v = 0 => 4 - 2t = 0

t = 2s.

- Draw distance-time graphs and corresponding velocity-time graph and acceleration-time graphs for the following:
- A ball dropped from the top of a tower
- A ball dropped from the top of a tower on to a perfectly elastic floor to be bounced back once
- A ball thrown upwards and then catch it again after a while
- The motion of an aircraft on landing after a long journey
- The motion of a parachutist when coming down
- The motion of a feather in the air

- The distance between two towns is 240km. Two cars move at 60km/h and 40km/h towards each other at the same time. Find when and where they meet up.
- Two trains remained stationary on the same railway track at a certain time, 60km apart. They moved towards each other at 40km/h and 20km/h respectively. At the same time, a bird started flying at 70km/h between the two trains up until they collided. How far would the bird have flown by the collision?
- A car moves at a constant speed of 20km/h. At the same time, a motorcyclist starts moving from rest, and accelerates at 5ms
^{-2}. Drawing velocity-time graphs on the same grid, find the following:- The time taken by them to have the same speed
- The distance travelled by each
- When and where they meet up

- A car starts from rest travels with an acceleration of 5ms
^{-2}for 4 hours. Then it slows down it velocity to 10km/h in the next two hours. During the last leg of the tour, the driver maintains the speed for 5 more hours. Draw a velocity time graph and find the total distance travelled. Hence, calculate the average speed for the journey.

Move the mouse over, just below this, to see the answers:

- Drawing
- 2.4hrs, 96km, 144km
- 70km
- 4hrs
- 40km, 80km
- 8hrs, 160km

- 120km, 10.9km/h

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7^{th} edition in print.