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dk bose:
1) A force acts on a particle,where R=(7i+16j)N.the force R is the resultant of two forces P and Q.the line of action of P is parallel to the vector (i+4j) and the line of action of Q is parallel to the vector (i+j).
Determine the forces P and Q expressing each in terms of i and j.

2)At 11:00 hours the position vector of an aircraft relative to an airport O is (200i+30j)km,i and j being unit vectors due east and due north respectively.The velocity of the aircraft is (180i-120j)km/h.find:
(a) how far it is from O at 12:00.

3) this question is quite huge so i cant write it here.it is from heinemann modular mechanics  unit 1 textbook.page no 57.review exercise 1.question no 7,part (f).

Solve all these 3 questions and give the solutions as fast as possible!!!!!

astarmathsandphysics:
For q3 I don't have the textbook. Can you scan it?

Becca:
Okay, I have solved the first two questions. I'm doing CIE A levels, so I don't have the book for the last question... and I couldn't find that particular page on the Internet. Drat it. Hope you understand my solutions! :)

First Question:

P is parallel to (i+4j), so P= x(i+4j)
Q is parallel to (i+j), so Q= y(i+j)

x & y are to be calulated, as follows:
since R is the resultant of P and Q, R = P+Q

therefore,
R = P+Q
(7i+16j) = x(i+4j) + y(i+j)
(7i+16j) = (xi + 4xj) + (yi +yj)
(7i+16j) = [(x+y)i + (4x+y)j]

Hence, based on the above equation, you can say that:
x+y = 7
y=7-x     >>>>>>>>>>>>>>>>> eqn 1

and,
4x+y = 16
y= 16-4x    >>>>>>>>>>>>>>>>>>>eqn 2

Sub eqn 1 into eqn 2:
7-x=16-4x
3x=9
x=3

Sub x=3 into eqn 1
y=7-3
y=4

Therefore, P=x(i+4j)
                P= 3(i+4j)
                P=3i + 12j

and Q=y(i+j)
      Q=4(i+j)
       Q=4i+4j


Second Question:

1 hour passes from 11.00 to 12.00, so the plane moves through a vector of (180i-120j) within this time. The initial vector is (200i+30j).
Therefore, the final vector is:

(200i+30j) + (180i-120j) = (380i-90j)

However, since the question is asking for the distance of the aeroplane from the airport, we must find the magnitude of the above vector via Pythagoras' theorem:

?[380^2+ (-90)^2] = 390.5 km

^For some reason, I couldn't insert the square root sign; it became a '?' instead.

Becca:
Oh no, I'm sorry, Mr Astarmathsandphysics! When I started writing the reply, you had not replied yet! Sorry again, I didn't notice that you had replied already before I posted mine...

astarmathsandphysics:
Is ok. If you had posted first I might have posted anyway.

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