Unit 23: Applied mathematics for engineering
LO3 and 5
Modelling the motion of a car
Instructions and answers for teachers
These instructions should accompany the OCR resource ‘Modelling the motion of a car’ activity which
supports Cambridge Technicals in Engineering Level 3.
Version 1
The Activity:
Learners are given a situation in which a car starts from rest and accelerates with a constant driving force.
Two different mathematical models are presented for which learners should solve and comment on their applicability.
The spreadsheet Car_motion_Data contains two sheets corresponding to the two mathematical models used.
It is recommended that learners be given time to complete this activity themselves after which the teacher could
present detailed solutions.
This activity offers an
This activity offers an
opportunity for English
opportunity for maths
skills development.
skills development.
!
Supporting
documents:
!
Formulae Booklet for Level 3 Cambridge Technicals in Engineering (available from
http://www.ocr.org.uk/qualifications/cambridge-technicals-engineering-level-3-certificate-extended-certificatefoundation-diploma-diploma-05822-05825/)
Spreadsheet Car_motion_Data
Suggested timings:
Task 1: 1 hour
Task 2: 2 hours
Task 3: 2 hours
Activity 1
In this activity, learners are given minimal model parameters and are required to produce a modified
differential equation to reflect the situation. Learners are then expected to solve the equation and
comment on the results. Learners should be reminded of Newton’s second law of motion (force = mass x
acceleration) and told how it relates directly to this exercise.
Solutions
(i)
m
dv
 D  Fi  Fr  Fd
dt
m  1500
Fi  mg sin 0  0
Fr  mg (0) cos 0  0
1500
dv
3
 9000  v
dt
2
Version 1
1000
dv
 6000  v
dt
(ii)
1000 dv
1
(6000  v) dt
1000 
1
dv   dt
(6000  v )
 1000ln(6000 v)  t  C
ln(6000  v) 
t C
t

 C1
1000 1000
6000  v  C2 e t / 1000
v  6000  C2 e t / 1000
Since v = 0 when t = 0, C2  6000
v  6000(1  e t / 1000 )
(iii)
8 / 1000
) = 47.8
When t = 8 v  6000(1  e
(iv)
The speed after 8 seconds is about 107 mph which quite reasonable for a car with a fairly powerful
engine. The problem with this model is the terminal speed. If the car continues to be driven with the
constant driving force, according to the model, it will approach a speed of 6000 ms–1. For this reason the
model is not reasonable and does not provide a practical solution.
Version 1
Activity 2
Learners are now required to repeat Activity 1 but with a different drag force equation.
Results of this task should be compared with the results of Activity 1.
Solutions
(i)
m
dv
 D  Fi  Fr  Fd
dt
m  1500
Fi  mg sin 0  0
Fr  mg (0) cos 0  0
1500
dv
3
 9000  v 2
dt
2
1000
dv
 6000  v 2
dt
(ii)
1000 dv
1
(6000  v 2 ) dt
1000
1
dv   dt
(6000  v 2 )
 6000  v 
  t  C (This integral appears in the list of standard formulae)
ln
2 6000  6000  v 
1000
 20 15  v 
  t C
ln
15  20 15  v 
25
 20 15  v 
15

ln
t  C1

25
 20 15  v 
 20 15  v 
at


 20 15  v   C 2 e


Version 1
where a 
15
25
20 15  v  (20 15  v)C 2 e at
v(1  C2 e at )  20 15(C2 e at  1)
20 15(C 2 e at  1)
Since v = 0 when t = 0, C 2  1
v
(C 2 e at  1)
v
20 15 (e at  1)
(e at  1)
v  20 15
(1  e  at )
(1  e at )
(iii)
When t = 8 v  20 15
(1  e 8t )
 42.7
(1  e 8t )
(iv)
The speed after 8 seconds is about 94 mph which is similar to the result using the previous model and is
quite reasonable. The terminal speed is now acceptable at about 173 mph which would not be
unreasonable for a high performance car.
Activity 3
Learners are encouraged to complete this activity using a spreadsheet.
Solution
See spreadsheet: Car_motion_Data with Cr = 0.05 and angle of incline = 15o.
Speed at t = 8s: v = 22.5 ms–1
Terminal speed: 122 mph
Version 1
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