Unit 23: Applied mathematics for engineering
LO3 and 5
Modelling the motion of a car
Learner activity sheet
The diagram below shows a car moving with speed v ms–1 along a smooth, straight section of road with
an incline of θ measured anticlockwise from the horizontal. D N is the driving force delivered by the
engine and F N is a force opposing the forward motion.
According to Newton’s second law of motion (force = mass x acceleration), the motion of the car which
has a mass m kg can be modelled by the following equation:
m
dv
 DF
dt
where t s is time.
F N comprises three essential components, Fi N (force due to the weight of the car acting down the
slope), Fr N (rolling resistance) and Fd N (aerodynamic drag). The model now becomes:
Version 1
m
dv
 D  Fi  Fr  Fd
dt
You are given:
Fi  mg sin 
Fr  mgCr cos 
where:
g ms–2 is acceleration due to gravity,
Cr is a constant rolling resistance coefficient.
For this exercise use g = 9.81.
In practice Fd is dependent upon the speed of the car, the shape and surface area of the car and the
density of air. For simplicity below we will model this as either Fd = Cd v or Fd = Cd v2 giving constant
numerical values for Cd.
Activity 1
Assume the following:
Fd 
3
v
2
Cr = 0
θ=0
m = 1500
D = 9000 (and remains constant)
(i)
Show that the motion of the car can be modelled by the following equation.
1000
dv
 6000  v
dt
Version 1
(ii)
The car starts from rest at time t = 0 and accelerates along the road.
Show that the speed of the car, v, at time t is given by:
v  6000(1  e t / 1000 )
(iii)
Calculate the speed of the car 8 seconds after it starts to accelerate.
(iv)
Comment of whether the model appears to represent a practical situation.
Activity 2
Use the same parameters as Activity 1 but this time use Fd 
3 2
v .
2
(i)
Show that the motion of the car can be modelled by the following equation.
1000
dv
 6000  v 2
dt
(ii)
The car starts from rest at time t = 0 and accelerates along the road.
Show that the speed of the car, v, at time t is given by:
v  20 15
where a 
(1  e  at )
(1  e at )
15
25
(iii)
Calculate the speed of the car 8 seconds after it starts to accelerate.
(iv)
Compare this model with the model used in task 1and comment on whether results obtained are
reasonable for this situation.
Version 1
Activity 3
Repeat Activity 2 with the same parameters but this time assume that the car is travelling up a gradient
with and angle of θ = 15o and with a constant rolling resistance coefficient Cr = 0.05.
You may find it helpful to enter values and equations into a spreadsheet to produce numerical results.
This will also allow you to modify parameters quickly and produce alternative solutions.
In addition you could create a table of values representing time and speed and use this to plot a
corresponding graph.
You could also include other calculations to approximate the distance travelled by the car for certain time
intervals.
Version 1