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noobie93
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#1
Report Thread starter 11 years ago
#1
Okay so here's the question:

A particle P of mass 2kg is moving under the action of a constant force F newtons. When t=0, it has velocity (3i+2j)ms-1 and at time t=4s, P has velocity (15i-4j)ms-1. Find

(a) the acceleration of P in terms of i and j
- I have done this, it is (3i-1.5j)ms-2

(b) the magnitude of F
- Also done this, it's 6.708N

(c) the velocity of P at time t=6s

Now my problem lies here. I thought that since velocity = acceleration x time, I could do 6 x (3i-1.5j) but I cannot, since that gives me an incorrect answer.

I then thought I could use SUVAT with t=6, u=(3i+2j), a=(3i-1.5j) to find v with v=u+at. However, again, this gave me an incorrect answer.

If it is possible could someone please tell me
(a) Why my methods were wrong and
(b) How to obtain the correct answer?

Thank you so much!
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roar558
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#2
Report 11 years ago
#2
v-u=at
v=at+u, you forgot to add the initial velocity
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04bellot
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#3
Report 11 years ago
#3
no its velocity is theoriginal vestor time added to the acceleration times the time :/
i.e (3i+2j) + (3i-1.5j)ms-2*6 seconds
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Insomniac_1901
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#4
Report 11 years ago
#4
i got (21i - 7j)m/s

have you got the answer?
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Insomniac_1901
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#5
Report 11 years ago
#5
using v = u + at

v = (3i + 2j) + 6*(3i - 1.5j)
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noobie93
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#6
Report Thread starter 11 years ago
#6
Ahh yes that works now, thanks! The answer is 21i-7j
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bradford9
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#7
Report 3 months ago
#7
how did u work out part b)
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Sinnoh
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#8
Report 3 months ago
#8
(Original post by bradford9)
how did u work out part b)
This thread is old enough to have done its year 6 SATs, if you clicked on a thread from the Trending or Related Discussions tab then check the date because it could be ancient.
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