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Differential Equations (videos, notes and quizzes with corrections) - Welcome To Class
Differential Equations (videos, notes and quizzes with corrections): Introduction Topics
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1 - Welcome To Class
: 0:03:44 mins
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Introduction
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Instructors
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MECHANICAL ENGINEER
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Ecole Normale Supérieure des Travaux Publics
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NATIONAL ADVANCED SCHOOL OF PUBLIC WORKS
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NATIONAL ADVANCED SCHOOL OF PUBLIC WORKS YAOUNDE
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Faculty of Engineering and Technology of the University of Buea
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Faculty of Engineering and Technology of the University of Buea
Please help me solve this problem
Question 2 june 2018 from your calculations e=1/3
Page 3
Page 2
Solution to June 2018 qn 7 by Johan
This is thé solution to the question (# Q6 PMM GCE jeune 2018)
Question 7 June 2018
Using : P/v = 14.28×1000/6m/s
=2371.6N
a- considering the car and the carriage as a single body the acceleration of the system
F - R = ma
2371.6 - (639 - 280) = ( 1400 +700)a
2021.6 = 2100a
a= 0.96m/s^2
b- T - 280 = 700a
T - 280 = 490
T = 770N
c- F = P/V
= 14.28 × 1000/12
= 1190N
So, F - R = ma
1190-(630-230) = 1400+ 700a
280 = 2100a
a = 0.13 m/s^2
d- time taken ,
We have first,280=700a
a=-0.4m/s^2
Retardation of its carriage at rest
Using
V= u + at
0= 12 + ( -0.4)t
-12= -0.4t
T= 30 s
Faith's solution
That was good Faith but you made a mistake when calculating the acceleration in the (a) part of the question. You were supposed to add the resistances not subtract them. F-R=ma, R = 639 + 280 = 919, a = (F-R)/m = (2371.6-919)/(1400+700) = 0.7 in meters per squared seconds
This is the solution to the question on impacts (# GCE june 2018 Q7)
You've not uploaded any solution Rene, please do so this evening