Dear Secondary Math students, Math Lobby will be going through Distance, Speed and Time (Kinematics) today. Kinematics will be tested in the form of real world context questions in O Level Elementary Math and N Level Math examinations.

Students are also required to learn how to draw both distance-time and speed-time graph. Without further ado, let's begin!

__In this note, you will learn:__

__1. The relationship between speed, distance and time and its formula__

__2. What does the gradients of distance-time and speed-time graph mean?__

__3. What does the area under a speed-time graph mean?__

__1. The relationship between speed, distance and time and its formula__

__1. The relationship between speed, distance and time and its formula__

Whenever we see a cheetah sprint on a documentary or a sports car racing by the streets, we will exclaim, “Wow, that’s fast!”.

However, this perception is subjective to the eyes of each individual, so how can we have a specific way of determining how “fast” or “slow” something is? This is where the measurement of how fast or slow something is comes in, or which what we call “**speed**”.

The speed of an object is the magnitude of the change of its position, and we typically measure it by taking the **total distance of an object travelled divided by the total time taken for the object to travel from the start to the end**. The formula is given by:

**Average Speed = Total Distance Travelled / Total Time Taken**

**(where the unit of average speed is usually given by meter per second, or m/s)**

Given that we have a cheetah that is running at its maximum speed of 93km/h, this basically means that assuming that the cheetah runs at a constant speed, for every hour that it runs, it would have covered a distance of 93km or 93000m, how amazing is that!

Another way to remember this formula is to memorize the **triangle method**. Looking at the diagram shown above, the horizontal line represents **division**, and the vertical line represents **multiplication**. This means that:

To get

**speed**, you must**divide distance**by**time**To get

**distance**, you must**multiply speed**by**time**To get

**time**, you must**divide distance**by**speed**

__2. What does the gradients of distance-time and speed-time graph mean?__

__2. What does the gradients of distance-time and speed-time graph mean?__

Now that we have got the basics of the relationship between speed, distance and time laid down, we will now be talking about the graphs of distance-time and speed-time graphs and what their gradients mean.

If you have difficulties understanding or finding the gradient of a graph, you can refer to Math Lobby’s article on “** How to find the gradient of a straight line**”.

From the diagram above, you can see that there are two lines, L1 and L2, on the distance-time graph. The gradient of a distance-time graph gives you the** speed** of an object between the two points you chose, and the **steeper** the gradient of the line, the **higher** the magnitude of the speed and likewise for the opposite, the **less steep** the gradient of a line on a distance-time graph, the **lower** the magnitude of the speed.

Similarly, you can see that there are two lines, L3 and L4, on the speed-time graph. The gradient of a speed-time graph gives you the** acceleration** of an object between the two points you chose, and the **steeper** the gradient of the line, the **higher** the magnitude of the acceleration and likewise for the opposite, the **less steep** the gradient of a line on a speed-time graph, the **lower** the magnitude of the acceleration.

From the diagram we see above, both the lines, L5 and L6, are shown on the distance-time and speed-time graphs respectively and both of them are increasing before turning into horizontal lines at t1.

Let’s firstly take a look at the distance-time graph. When a line turns horizontal on a distance-time graph, it means that regardless of the time taken, the distance still remains the same/ is constant, what does this mean?

This means that the line has a gradient of **zero**, and it is **no longer in motion** after t1! Because the distance is no longer increasing/decrease, which shows that the object is at rest.

Next, let’s take a look at the speed-time graph. When a line turns horizontal on a speed-time graph, what does it mean? Leave your answer in the comments down below before viewing the answer!

This means that the line on the speed-time graph has a gradient of **zero**, and **speed has become constant** after t1! It is different from the distance-time graph as a horizontal line on a speed-time graph does not necessary mean that the object stopped moving, but it is travelling at the same speed for the specific time period. (**Unless if s = 0, then yes, the object has stopped moving and is at rest.**)

__3. What does the area under a speed-time graph mean?__

__3. What does the area under a speed-time graph mean?__

Now that you have understood the concepts of what gradients represent in a distance-time graph and a speed-time graph, we will now look at the area under a speed-time graph.

Area under a speed-time graph basically represents the **total distance travelled**. Let’s take a look at the example below:

Given that a car is travelling on a road, the journey is depicted in the graph of the diagram above. The car travels at an increasing speed until it reaches t1, and it then it travels at a constant speed until it reaches t2, which he proceeds to slow down his car and comes to a stop. Given that t1 = 5s, t2 = 15s and s1 = 20m/s, find the total distance travelled by the car from t = 0s to t = 15s.

To solve this, we will have to break the graph into parts: t = 0s to t = 5s, and t = 5s to t = 15s.

Therefore, **total distance travelled by the car from t = 0s to t = 15s**

**= [½ x (t1 – t0) x s1] + [s1 x (t2 – t1)]**

= [½ x (5 – 0) x 20] + [20 x (15 – 5)]

= (½ x 5 x 20) + (20 x 10)

= 50 + 200

= 250m

The total distance travelled by the car from t = 0s to t = 15s is 250m.

__Question Time!__

1.. The diagram above shows the speed-time graph for a train journey.

a) Calculate the deceleration of the train for the last 10 seconds of the journey

b) Calculate the total distance travelled on the journey

c) The maximum of speed of the train was 15 m/s, change 15 m/s to km/h.

2. John runs at a constant speed from home to a mall. After spending 10 minutes in the mall, he walks back home at a constant speed. He is away from home for 30 minutes.

The first part of his journey is shown on the distance-time graph below.

a) On the grid, complete the graph for Ali’s journey.

b) Calculate John’s speed as he runs to the mall. Give your answer in kilometers per hour.

c) On the grid below, draw the speed-time graph for the whole of Ali’s journey.

3. The diagram shows the speed-time graph for a car’s journey between two sets of traffic lights. The shaded area represents the distance travelled. The distance travelled is 579m.

a) Calculate the greatest speed, v m/s, of the car.

b) Calculate the acceleration for the first 37 seconds of the journey.
And that’s all we have for today! ** Math Lobby** hopes that after reading this article, you have a clear understanding of the relationship between speed, distance and time, know the meanings of the gradients of a distance-time and speed-time graph, and what the area under the speed-time graph represent.

As always: Work hard, stay motivated and I wish all students a successful and enjoyable journey with Math Lobby!

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