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Old May 21st 2009, 11:24 AM   #1
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Question Question on vectors

Please can you help me out with these questionsefine the foollowing terms with two examples each:A null vectors,A unit vectors,and collinear vectors.

Find the direction cosines(L,m,n) of the vector p=3i-2j+6k.

Suppose the motion of a particle is described as
S=a+b/2t2
Where a=50cm and b=10cm s2
Find: the displacement of the particle in the time interval between
T1=2s and t2=3s
the instantaneuous velocity at time t1=4s

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Old May 21st 2009, 02:45 PM   #2
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Hi, I hope this will help you to get started. A null vector is just A zero

vector e.g (0,0) in R^2 and (0,0,0) in R^3 (R denotes the real numbers).

A unit vector is a vector of norm e.g (1,0,0) and (0,1,0) both in R^3.

Collinear vectors are vectors which are non-zero scalar multiples of each other.

Example1 (1,1,1), (2,2,2) and example 2, (2,3,4), (4,6,8)

of each other.

||p||=$\displaystyle \sqrt{3^2 + 2^2 + 6^2}$= 7.

Hence L=3/7, M=-2/7 and N=6/7 (assuming L represents the x axis, M the y axis and N the z axis).

For the next bit of the question I would assume constant acceleration.

Last edited by C.E; May 22nd 2009 at 05:28 PM.
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Old May 22nd 2009, 12:05 PM   #3
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Post Appreciation

Thanks for your explanation and am really appreciate that ,but,l was confused about what you said,quote(for the next bit of the question l would assume constant acceleration) Please explain further.And am waiting for you reply on the next question
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Old May 23rd 2009, 04:41 PM   #4
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I was a little confused about what you were asking for the second part of the question. By the following:
Originally Posted by Ralph4christ View Post
S=a+b/2t2
Do you mean that $\displaystyle S=\frac{(a+b)}{2}t$ ?

Also do you mean that t1=2 and t2=3 (with units of seconds)?
Originally Posted by Ralph4christ View Post
T1=2s and t2=3s
Also when you say "a=50cm and b=10cm s2", what exactly do you mean?

The reason I said to assume constant acceleration was that $\displaystyle S=\frac{(a+b)}{2}t$ is a standard formula for an object undergoing constant acceleration (if a= initial velocity and b=final velocity).
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Old May 24th 2009, 11:28 AM   #5
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Post Explanation

Thanks very much for your effort all tihs while,anyway what l means here is exactly what you get,Soppose the motion of a particle is described as S=(a+b)/2, t square.Also l mean t underscore 1 =2seconds and t underscore 2 =3seconds.

And where a =50cm and b=10cm per seconds square
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Old May 29th 2009, 03:43 AM   #6
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Enquiries

Please hope there's no problem for not hear from you all this while,anyway l believed it well and am still looking forward to hear from you,On the remaining question. thanks
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Old May 30th 2009, 04:01 PM   #7
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Sorry but, I am still a little confused about what you are asking
And where a =50cm and b=10cm per seconds square
Do you mean a and b are in $\displaystyle ms^{\ -1}$ ?

Also what does T1 stand for?

When you say
Find: the displacement of the particle in the time interval between
do you mean find the change in the displacement of the particle between t=2 and t=3 seconds?

Last edited by C.E; May 30th 2009 at 05:26 PM.
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Old May 31st 2009, 11:43 AM   #8
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Post Explanation

Yes l mean find the change in the displacement of the particle between t1=2secs and t2=3secs.

.

S=a+b/2tsquare, a=50cm and b =10cmper square

S1=50cm+10cm per secs square*2ssquare
50cm+5cm per secs square*4secs
=50cm+20cm
S1=70cm.

S2=50cm+10cm/2*3secs square
=50cm+5cm per secs square*9secs
50cm+45cm
S2=95cm.

Therefore, Displacement of the particle in the time interval between S1=2secs and S2=3secs.
=S2-S1=(95-70)CM
=25CM.

To find the instantaneous velocity at t1=4secs
V=change in displacement/change in time

25cm/4secs=5m per secs square

Anyway this is how l get it,please tell me if am right
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