# Bar Model Method for Speed Problems – A ratio Approach

##### Ho Soo Thong & Ho Shuyuan

First Edition ,Nov 2014, First Revised Edition, July 2016

Abstract
This article illustrates the use of Bar Model Method to visualise and relate the simple algebraic relation in speed problems for effective ratio approach.

## Simple Speed Problems

In our earlier work, the term simple journey of A means a journey in which A travels at the same speed throughout when A is moving. The term travelling time of A means the time taken by A only when A is moving (See [1] and [2]). We will solve problems involving two simple journeys, applying the algebraic relation

distance = speed × time

There are many challenging simple journey speed problems at Primary level and at Primary Olympiad level. An example is the well known puzzle The fly and the bicycle problem by Martin Gardner (1914 – 2010).

The fly and the bicycle problem

In Example 1, we begin with a variant of this problem.

## Simple Journeys with Same Travelling Time

### Example 1

Two cyclists A and B, 30 km apart, moving towards each other at a speed of 10 km/h and 20 km/h respectively. A fly flies between the handles of the bicycles at a speed of 15 km/h until they meet.
What is the total distance travelled by the fly?

### Solution

A ratio approach to the problem is shown below.
Noting that they started at the same time and ended together, they have the same travelling time. So the speed ratio and the distance ratio for each pair are the same and so the distance travelled by the fly is 15 km as shown in Figure 1.

The above example illustrates a general result shown in Figure 2.
Figure 2 shows two simple journeys with same travelling time and

speed ratio = distance ratio

Next, we will proceed with an another ratio feature.

## Simple Journeys with Same Speed.

Figure 3 shows two simple journeys in two consecutive distances and the two corresponding consecutive travelling times. Then

time ratio = distance ratio

The next example will see the feature in application.

### Example 2

At 08 30, Steve started to walk from home to school at a certain speed throughout.
At 09 45, his brother Tom left home for the same school at a certain speed throughout on a bicycle and he reached school at 10.45.
At 10.30 both of them passed a police station that was 4 km from school .
(a) At what time did Tom reach the school?
(b) What was the distance between their home and school?
(c) What was Steve’s walking speed? What was Tom’s walking speed?

Solution
Let
(a) In Figure 4, the bar models depict two consecutive simple journeys for Tom and Steve:

H (Home) to P (Police station)

and then

P to S (School).

For Tom, the time ratio is

travelling time ( H → P ) : travelling time ( P → S ) = 45 : 15 = 3 : 1

For Steve

travelling time ( H → P ) = 3×40 (= 120) minutes,

and so

travelling time ( P → S ) = 40 minutes.

Steve reached the school at 11 10 as shown.

(b) In the third bar model, the distances travelled from H to P and then from P to S are in the ratio 3 : 1. Since PS = 4 km, then HP = 12 km and therefore the distance between their home and school is 16 km.

(c) In the bar model, Steve walked 12 km in 2 hours (H to P)., i.e. Steve’s speed is 6 km/h. Tom cycled 16 km in 1 hour (H to S), Tom’s speed was 16 km/h.

Noting that Tom traveled at a faster speed of 16 km/h and therefore he had a shorter travelling time of 60 minutes; Steve travelled at a slower speed of 6 km/h and therefore he had a shorter travelling time of 160 minutes; this leads us to the following ratio feature.

## Simple Journeys with same distance travelled.

For two simple journeys on the same road for the same distance travelled, For the journey with a the faster speed u, the shorter travelling time is a. For the journey with slower speed v, the longer travelling time is b as shown in Figure 6.

This result will be applied in the next two examples.

### Example 3

At 7 am, Sue walked from home to school at a speed of 6 km/h throughout. She was 8 minutes late.
The next day, she cycled from home to school at a speed of 10 km/h throughout. She was 12 minutes early.
(a)What is the latest reporting time for her school?
(b)What is the distance between her home and school?

### Solution

Since

Walking speed : Cycling speed = 6 :10 = 3 : 5,

then

Walking time : Cycling time = 5 : 3.

For the scenarios of being late and early, we construct a yellow bar model for the latest reporting time and then we construct two time bar models for the time ratio 5 : 3.

(a)The latest reporting time is 7.42 am.
(b) For the required distance, we consider the second day trip and he travelled for 30 minutes for a distance of 10 × 1/2 km = 5 km.

### Example 4

Two cars A and B travel at constant speeds from X to Y which are 180 km apart. The car B travels at a speed 20% slower than the speed of A and so it will take B 20 minutes more than A.
(a) What is the time taken by A?
(b) What the speed of B?

### Solution

(a) Noting that 20% = 1/5, Figure 8 shows the bar model for the speed ratio

Speed B : Speed of A = 4 : 5

Figure 9 shows of time ratio

Travelling time of A : Travelling time of B = 4 : 5

In Figure 9, each unit is 20 min and the total time taken by A is

4 units = 4×20 min = 1h 20 min.

(b) In Figure 10, we construct a distance bar model for the Car B with 5 identical bars corresponding time model of B with 5 identical bars. Each bar represents the distance of 36 km travelled in 20 minutes. The distance travelled in 1 hour (60 min ) is 108 km.
and so the speed of B is 108 km/h.

Remarks
A problem solving approach is to identify simple journeys, relate simple journeys and apply ratio approach. There are many challenging speed problems can be solved with this approach (See [ 1],[2 ])

References
[1] Ho Soo Thong, Ho Shuyuan, Speed Problems – A Ratio Approach, Mathematical Medley, Vol 37, No. 1-2, Dec 2011.
[2] Ho Soo Thong, Ho Shuyuan and Leong Yu Kiang Problem Solving methods for Primary Olympiad Mathematics, AceMath, 2012.