Since the graph is a linear, the equation y=mx+c can be used to model its behavior where m stands for the gradient and c is the y-intercept.
Using this linear equation, we have to find m:
m=x2- x1
y2-y1
by substituting theses values to find the gradient from these two co-ordinate points (32,0.006) and (48,0.009):
m=0.009- 0.006 =1.875 x 10^-4
48-32
the y-intercept of the linear is 0 because as the speed is 0 there is no need to think to brake the vehicle so:
c= 0
the final function we get from using the equation is:
y=1.875*10 ^-4x+0
Since there cannot be any negative thinking distances or speed, the equation for the linear of speed versus thinking distance has to be:
y=|1.875*10 ^-4x+0|
Speed versus Braking distance
From the values of the braking distance, there isn’t a common difference and the braking distance rising exponentially creates a curve suggesting a parabola.
Therefore the quadratic equation, y=a(x+ a)² also known as y=ax²+bx+c can be applied to model its behavior.
As the speed is 0 the braking distance would be 0 as well therefore the y intercept will be a repeated root, this will mean that a=0 given the equation y=a(x+a) ²
y=a(x+ a)²
=a(x+0)²
=a(x)²
y=ax²
To solve y=ax², the co-ordinates (48,0.014) from the graph can be used to be substituted into the graph:
y=ax²
0.006=a(32)²
0.006=a(1024)
a=0.006
1024
a=5.859375*10^-6
Quadratic equation of Speed versus Braking distance:
y=5.859375*10^-6 x²
Like the linear Speed versus Thinking graph above, there cannot be negative speeds, thinking distances or braking distances therefore the equation has to be:
y=|5.859375*10^-6 x²|
By adding the thinking and braking distance added together the overall average stopping distance can be obtained. Below is the graph of each of the distances versus the speed:
Speed versus: Thinking distance
Braking distance
Overall average stopping distance
The overall average stopping distance line, shown above, displays a similar parabolic curve like the braking distance with an exponentially growing y-value. This is in contrast to the linear of the thinking distance which is increasing in distance as well but at a slower rate.
Given that the overall average stopping distance share the same characteristic as the braking distance, the quadratic equation used above can be used here too:
As the speed is 0 the braking distance would be 0 as well therefore the y intercept will be a repeated root, this will mean that a=0 given the equation y=a(x+a)²
y=a(x+ a)²
=a(x+0)²
=a(x)²
y=ax²
The points (80, 0.053) are chosen to substitute into the equation:
y=ax²
0.053=a(80)²
0.053=a(6400)
a=0.053
6400
a=8.28125*10^-6
Therefore the equation for Speed versus Overall Average Stopping distance is:
y=8.28125*10^-6x²
For the reason that the overall average stopping distance can be found by adding the thinking and braking distance together, the functions of the Speed versus Thinking distance and Speed versus Braking distance can be added together to give the equation of the Speed versus the Overall Average Stopping distance.
y=function1+function2
Speed vs. Overall Average Stopping distance= (|1.875*10 ^-4x+0|) + (|5.859375*10^-6 x²|)
Here is another set of data which can be used to see if the model fits
Speed versus Stopping distance
As we can see, the points are forming a curve which is alike the previous quadratic graphs however, the equation of the graph:
May still be a quadratic equation but it is not exactly the same as the other stopping distance equation because different factors such as the tire friction, the weather/conditions, the driver’s age and the road’s surface all affect the overall stopping distance. Anomalies have to be taken into account as well as accidents can happen which affect the outcome. With more data, the accuracy of the graph can be improved so that the function can also be modified and become more precise.