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vectors and mechanics

Please can someone help me I'm confused with the first part:

At time t seconds a particle P has position vector r metres, with respect to a fixed origin O, where:

r = (4t^2 - kt + 5) i + (4t^3 + 2kt^2 - 8) j, t 0

When t = 2, P is moving parallel to the vector i.
(i) show that k = -5 (4 marks)

I get that the i vector for t and the original equation are the same, but I just don't get how you get an r value, or if you're even supposed to get an r value. If someone could please help I would really appreciate it - thank you in advance.
Original post by jess.harrison
Please can someone help me I'm confused with the first part:

At time t seconds a particle P has position vector r metres, with respect to a fixed origin O, where:

r = (4t^2 - kt + 5) i + (4t^3 + 2kt^2 - 8) j, t 0

When t = 2, P is moving parallel to the vector i.
(i) show that k = -5 (4 marks)

I get that the i vector for t and the original equation are the same, but I just don't get how you get an r value, or if you're even supposed to get an r value. If someone could please help I would really appreciate it - thank you in advance.


'moving parallel to the vector i' - what do you think this mean?
Original post by Muttley79
'moving parallel to the vector i' - what do you think this mean?

That the motion of P has the same direction as i in the equation? I'm not too sure how to word it though...
Original post by jess.harrison
Please can someone help me I'm confused with the first part:

At time t seconds a particle P has position vector r metres, with respect to a fixed origin O, where:

r = (4t^2 - kt + 5) i + (4t^3 + 2kt^2 - 8) j, t 0

When t = 2, P is moving parallel to the vector i.
(i) show that k = -5 (4 marks)

I get that the i vector for t and the original equation are the same, but I just don't get how you get an r value, or if you're even supposed to get an r value. If someone could please help I would really appreciate it - thank you in advance.


No, you don't need an "r" value for this first part.

Moving parallel to the vector i, means it's velocity is parallel to the vector i. And if it's velocity is parallel to the vector i, what's the j component of the velocity?
(edited 5 years ago)
Oooooohhhhhh I see, thank you I'm being stupid :smile:
Original post by ghostwalker
No, you don't need an "r" value for this first part.

Moving parallel to the vector i, means it's velocity is parallel to the vector i. And if it's velocity is paralllel to the vector i, what's the j component of the velocity?
Reply 5
Original post by jess.harrison
Please can someone help me I'm confused with the first part:

At time t seconds a particle P has position vector r metres, with respect to a fixed origin O, where:

r = (4t^2 - kt + 5) i + (4t^3 + 2kt^2 - 8) j, t 0

When t = 2, P is moving parallel to the vector i.
(i) show that k = -5 (4 marks)

I get that the i vector for t and the original equation are the same, but I just don't get how you get an r value, or if you're even supposed to get an r value. If someone could please help I would really appreciate it - thank you in advance.


Hi, where did you get this question from?
Reply 6
Original post by Hdbfjdbdn
Hi, where did you get this question from?

This thread is 4 years old and that poster hasn't been online since 2019 :smile:

Is there a specific reason why you need to know where the question comes from - it looks like a fairly standard vector mechanics question (i.e. I've seen several similar posts on TSR over the past few years)?
Reply 7
Hi,
Oh I see, well yes I know it's an OCR question as I saw it in a question bank but wanted to know the specific paper as I wanted to do it. I searched on Google but couldn't find the exact paper.
Reply 8
Original post by davros
This thread is 4 years old and that poster hasn't been online since 2019 :smile:

Is there a specific reason why you need to know where the question comes from - it looks like a fairly standard vector mechanics question (i.e. I've seen several similar posts on TSR over the past few years)?

Have you got any idea where it could be from?
Reply 9
Original post by Hdbfjdbdn
Have you got any idea where it could be from?

No, but you can still attempt the bit that's posted originally with the hints given (Looking at it I think there may be a small typo in the OP but not sure - if it's in a question bank you've seen, perhaps you could post the complete question with your attempt and then someone can check it for you :smile: )
(edited 9 months ago)

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