Typical stopping distances

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
d
; B_d = Braking distance, B
d
; 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
d
T
against
v
, where
d
T
is the thinking distance in feet and
v
is
the speed in mph.
Solution: A table of speed v(mph) vs T
d
(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
d
and v and plot to obtain the graph
shown below from Ms-excel.
Question 2: Find an equation that connects
d
T
and
v
.
Solution: From the graph, the thinking distance and speed exhibit a relationship
which shows a direct proportionality. That is,
d
T
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 .
0
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
v
. 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
d
B
and
v
is thought to be quadratic,
0
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
45
d
B
then
30v
and
125
d
B
then
50v
, find the values of a and b and hence write down the equation connecting
d
B
and
v
.
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,


  

 
2
d
B av bv c
Surname 5
Question 6: Write down an equation which connects the overall stopping distance,
s
to the
speed
v
.
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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