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# There are three snails; slippery, slimy and slidey. They enter a ten-metre race for food. Each snail runs according to the following rules. Slippery : d= 4.4 + 0.55t Slimy : d= 0.3t(t-7) Slidey : d= 0.3t(t-3.4)(t-9)

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Introduction

(Speedy Snails) ## INTRODUCTION

Mathematics can be used to solve the problem happened in our life such as finding distance or time. Now, we’ve got a problem here. We want to know following questions below. Let us solve the problem by using mathematical method.

There are three snails; slippery, slimy and slidey. They enter a ten-metre race for food. Each snail runs according to the following rules.

##### Slippery        :        d= 4.4 + 0.55t

Slimy                :        d= 0.3t(t-7)

Slidey                :        d= 0.3t(t-3.4)(t-9)

The snails race from a designated starting point toward a designated finish line. The distance, d, is measured in metres, and the time, t, is measured in minutes.

## QUESTIONS

1.        Find the distance of each snail from the start after:    (An appropriate window and graph)

There are two ways to solve the question 1.

Firstly, we can substitute the time into each formula.

1. 0 minutes

Slippery         :         d = 4.4 + 0.55t

= 4.4 + 0.55(0)

= 4.4 + 0

= 4.4(m)

Slimy                 :         d = 0.3t(t-7)

= 0.3(0)(0-7)

= 0(-7)

= 0(m)

Slidey                 :          d = 0.3t(t-3.4)(t-9)

= 0.3(0)(0-3.4)(0-9)

= 0(-3.4)(-9)

= 0(m)

### Name of snails

Slippery

Slimy

Slidey

Distance (m)

4.4

0

0

(A distance of each snail when the time is 0 min.)

1. 2minutes

Slippery         :          d = 4.4 + 0.55t

= 4.4 + 0.55(2)

= 4.4 + 1.1

= 5.5(m)

Slimy                 :          d = 0.3t(t-7)

= 0.3(2)(2-7)

= 0.3(-10)

= -3(m) (This means Slimy is going backwards)

Slidey                 :          d = 0.3t(t-3.4)(t-9)

= 0.3(2)(2-3.4)(2-9)

= 0.3(19.6)

= 5.88(m)

 Name of snails Slippery Slimy Slidey Distance (m) 5.5 -3 5.88

Middle

Distance (m)

7.15

-3

-9.6

(A distance of each snail when the time is 5min.)

State which snail is leading the race at each time

According to the results above

@ Slippery         is leading the race at 0 min.

@ Slidey         is leading the race at 2 min.

@ Slippery _         is leading the race at 5 min.

Secondly, we can use a graphics calculator. If we use a graphics calculator we would know not only the distance of each snail at each time but also which snail is leading the race at each time very simply and easily by comparing 3 points on the graph.

<Slippery>                   <Slimy>                  <Slidey>   @(Slippery is leading the race at 0 min.)

<Slippery>                     <Slimy>                           <Slidey>   @(Slidey is leading the race at 2 min.)

<Slippery>                      <Slimy>                         <Slidey>   @(Slippery is leading the race at 5 min.)

Conclusion:

As you can see graphs above

@ Slippery is leading the race at 0 min.

@ Slidey is leading the race at 2 min.

@ Slippery is leading the race at 5 min.

1. With the aid of your graphical calculator, sketch the path of each snail include each graph in your report. < y = 4.4 + 0.55t > < y = 0.3t(t-7) > < y = 0.3t(t-3.4)(t-9) >

3.        At what time(s) is each snail at the starting point.

Conclusion

2 + 9t +28) Slippery will go 237(m) in 6 minutes.

We could also find the time taken by snail by using graphics calculator.

Example:

How long will Slimy take to complete 17 metre race?

(But, Slimy runs according to the following rules. d= 0.8t+16)  Slimy will take 1.25 (min.) to complete 17 metre race.

Conclusion:

There are 2 methods that can always be used to predict the position of snail at any time and the time taken by each snail to reach any position.

Substitute the time (the distance) in a function given.

use a graphics calculator for accuracy and convenience

7.        Generalize your methods in question 6 to solve a function d(t) for any distance, d, or, time, t

I tried to take another kind of d(t) function to generalize my methods in question 6.

d(t) = (3/x)+5                       d(t) =e2t-2

d(t) = (3/x) + 5

By using first method.

1. Let us suppose t = 9(min.)

Substitute t = 9(min.) in function

d         = (3/9) + 5

d          = 5.33(2.d.p)

in this function, distance(d) = 5.33(2.d.p) when the time(t) = 9(min.)

2. Let us suppose d = 8(m)

Substitute d= 8(m) in function.

8         = (3/x) + 5

8-5         = (3/x)

3        = 3/x

x         = 1

the time(t) = 1 when the distance(d) = 8(m)

d(t) =e2t-2

By using second method (graphics calculator)

1. Let us suppose that time(t) = 2(min.) 2. Let us suppose the distance(d) = 5.5(m) Conclusion:

I’ve generalized that my two methods ( substitute the time or (the distance) use a graphics calculator) can always be used to find any distance, d, or, time, t.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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