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# portfolio Braking distance of cars

Extracts from this document...

Introduction

Mathematics                                                                                                                      000065022

Mr. Hosington                                                                                                               Kotaro Himi

1. Use a GDC or graphing software to create data plot of speed versus thinking distance and data plot of speed versus braking distance. Describe your results.

The data plot of speed versus thinking distance. The speed is relative to the thinking distance. So, this graph shows the direct proportion. There is no way that these values are minus because it doesn’t make sense that thinking distance is minus.

In this report, let the time of stepping on the brake is equal.

The data plot of speed versus braking distance. The graph should show exponential function. It is conceivable that it causes friction when the car breaks. When the car starts breaking, the speed of car is on the table below. During the car keep breaking, speed is decreasing due to friction.

 Speed (km/h) Speed(m/s) 0 0 32 8.888 48 13.33 64 17.77 80 22.22 96 26.66 112 31.11

The relationship between the speed and advanced distance is shown the graph below. x = speed (m/s) y=advanced distance First, the speed of car is constant. Speed will be decreasing according to braking cause friction. Thus the advanced distance will be decreasing as well. At least, the car will stop which indicate point 0. Therefore, area of triangle represents braking distance (BD).

Middle

a

a= = (0.00593)

・When (x=96,y=55)

55= a

a= = (0.00596)

・When (x=112,y=75)

75= a

a= = (0.00597)

Thus the average value of a is able to calculate

Average of a =      =0.005998

Consequently, the function will be

y=0.005998 Now, it is absurdly correct function which is able to show graph below. 1. The overall stopping distance is obtained from adding the thinking distance to the braking distance. Create a data table of speed and overall stopping distance. Graph this data and describe the results.
 Speed (km/h) Stopping distance (m) 32 12 48 23 64 36 80 53 96 73 112 96

Stopping distance is thinking distance add to breaking distance. x= speed y=stopping distance

The graph is exponential function because overall stopping distance is that thinking distance add to braking distance that contain the aspect of friction. Tthis graph which is related between speed and overall distance is more sharp than  other graph which is related speed and braking distance.

The connection with braking distance and thinking distance is able to show on the graph below. Thinking distance is direct propotion. Braking distance is exponential function. And, overall stopping distance will be exponential function because stopping distance represent that thinking distance added to breaking distance which shows above. So, the function of overall stopping distance should be the curve which can draw through top of the bar graph.

Conclusion

Thinking distance

y = Braking distance

y=0.005998 （ｙ＝ ｘ）＋（y=0.005998 = {2y=0.005998 + x}

= (y= 0.01199 +0.375x) That is certain wrong because that curve does not through any point. So the function which has value of a, b, and c is close to the appropriate curve.

1. Overall stopping distance for other speeds and given below. Discuss how your model fits this data, and what modifications might be necessary.
 Speed (km/h) Stopping Distance (m) 10 2.5 40 17 90 65 160 180 x= speed ,y=stipping distance

This graph shows exportional function as well.  Therefore it is possible to find the function according to use [y= +bx+c].

2.5= a+10b+c…①

17= a+40b+c…②

65= a+90b+c…③

②－① （17= a+40b+c）－（2.5= a+10b+ｃ）

＝{(17-2.5)=( - )a+(40-10)b}

=(14.5= 1500a+30b)…④

③－② （65= a+90b+c）― (17= a+40b+c)

= {(65-17) = ( - ) a+ (90-40) b}

=(48=6500a+ 50b)…⑤

⑤－④ (48=6500a+ 50b)-( 14.5= 1500a+30b)

= {(144-72.5)=(19500-7500)a+(150-150)c}

= (71.5=12000a)

= (a= ＝0.00595)

Substitute a value to ① and ②

2.5= × +10b+c…⑥

17= × +40b+c…⑦

⑦－⑥ (17= × +40b+c)- (2.5= × +10b+c)

=[(17-2.5)= {( × )-( × )}+ (40-10)b}]

= (14.5= (9.53333-0.59583) +30b

= (30b=14.5- 8.9375)

= (b= =0.18541)

Substitute value of a and b into the ①

2.5= × +10× +ｃ

ｃ＝2.5-(0.59583+1.8541)

ｃ＝2.5-2.44993=0.05007

Consequently, the function will be

y=  + x+0.05007

y=0.00595 +0.18541x+0.05007

This function is fits the data table below.

 Speed (km/h) Stopping Distance (m) 10 2.5 40 17 90 65 160 180 Moreover, this graph is fit other data as well.

 Speed (km/h) Stopping distance (m) 32 12 48 23 64 36 80 53 96 73 112 96 This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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