Stopping Distances Analysis

Surname 1

Name

Professor

Course

Date

TYPICAL STOPPING DISTANCES

Problem: From the information provided we obtain the table below.

In the table, T_d = the thinking distance T

; B_d = Braking distance, B

; C_l (m) = car

length in meters; C-L (ft.) = car length in feet; C_l (cars) = car length in terms of the number

of cars. The speed in meter-per-hour is v

1 mile = 1.0693 kilometers

1 mile = 5280 feet

Question 1: Plot a graph of

against

, where

is the thinking distance in feet and

the speed in mph.

Solution: A table of speed v(mph) vs T

(ft) is as shown below:

1 mile = 1.0693 kilometeres(km)

1 mile = 5280 feet (ft). Hence  





  The table becomes:

Surname 2

From the table above, we deduce the values of T

and v and plot to obtain the graph

shown below from Ms-excel.

Question 2: Find an equation that connects

and

Solution: From the graph, the thinking distance and speed exhibit a relationship

which shows a direct proportionality. That is,

is directly proportional to v. This

relationship is mathematically expressed as follows:





 . Removing the proportionality symbol and introducing the constant of

proportionality, we have:





 , where k is the constant of proportionality, equivalent to the average

thinking time .

20000

40000

60000

80000

100000

120000

0 10 20 30 40 50 60 70 80

Thinking distance, T_d

Speed, v

A graph of Thinking Distance Against

speed

Surname 3

Question 3: Using the information given, draw a graph of

against

. From your

graph, determine the braking distance when the car is 52 mph. If the breaking distance is 200

feet, what is the speed of the car?

Solution: We deduce the following table from the information given:

From the table we plot the values to obtain the graph shown below

The breaking distance when v=52mph, is 200,000ft

If the breaking distance is 200 ft, the speed is approximately equal to zero.

Question 4: The equation which connects

and

is thought to be quadratic,

50000

100000

150000

200000

250000

300000

350000

400000

0 10 20 30 40 50 60 70 80

Breaking distance, B_d (ft)

Speed, v (mph)

A graph of breaking distance vs Speed

Surname 4

What is the breaking distance when the car is stationary? What does this tell us about the

value of the constant c? Using the fact that when

B 

then

30v 

and

125

B 

then

50v 

, find the values of a and b and hence write down the equation connecting

and

Solutions: When the car is stationary, v=0, hence







 









 



   .

The breaking distance is a constant when the car is not moving. This means that the breaking

distance is partly constant.

Substituting the given values: 



 









 



 . This implies that:

       (i)

Also;  



 









 



 .This implies that:

      (ii)

Eliminating c in each case,

    (iii)

Since the initial breaking distance is 6km, from equation (i) we obtain:

     (iv)

Solving (iii) and (iv) simultaneously, 





  





. The equation is therefore

expressed as follows: 



 















  .

Question 5: Using the information given, check that that your equation works for the

braking distance given for a speed of 70 mph.

Solution: For the given speed, 



 





  









    

B av bv c  

Surname 5

Question 6: Write down an equation which connects the overall stopping distance,

to the

speed

Solution: The overall stopping distance   



 



   















 

Question 7: On a certain 2 mile stretch of road the police estimate that there are 50

cars traveling in the same direction. No overtaking is allowed on this stretch of road.

Investigate the maximum speed at which the cars can safely travel in both dry and wet

conditions.

Solution:       ,       

       















 

    















 . Working out and simplifying, the maximum speed:





 

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Stopping Distances analysis

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