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M1 vectors help!

Thanks for any help with the last question:

I'm trying a different one and I'm stuck on part c. will be grateful for any help.

Two cyclists, C and D are travelling with constant velocities (5i-2j) ms-1 and 8jms-1 respectively relative to a fixed origin O.

a) Find the velocity of C relative to D ( Found the answer - 5i - 10j )

At noon the position vectors of C and D are (100i +300j)m and (150i+100Jj) respectively, referred to O. At t seconds after noon, the position vector of C relative to D is s metres.

b) Show that s=(-50+5t)i +(200-10t)j ( I can do this question too )

c) By considering /s/^2 or otherwise, find the value of t for which C and D are closest together.

I'm not sure how to do this one, but I made a quadratic equation of t^2 - 30t - 20 = 0

But I solved this and this isn't the correct answer, where do I go from here?

Thanks in advance
(edited 11 years ago)
Reply 1
Original post by L1000
Hey, struggling on vectors, any help with this question will be appreciated very much!

I can do question a, but stuck on part b?

At time t=0, two ice skaters John(J) and Norma(n) have position vectors 40j and 20i metres relative to an origin O at the centre of an ice rink, where i and j are unit vectors perpendicular to each other. John has constant velocity 5im/s and Norma has constant velocity (3i +4j)m/s

a) show that the skaters will collide, and find the time at which the collision takes place.

b) on another occasion, John has position vector 40j metres. He wishes to skate in a straight line to the point with position vector 30i metres. Given that his speed is contant at 5m/s, find his velocity.

I know the equation r = r(position vector) + vt

So I sub in the numbers: 30i = 40j + vt?

I'm not sure on where to go from there, I'm thinking something with pythag since its speed, but sort of confused...

Thanks in advance


You need a vector from 40j to 30i with magnitude 5. Looks like Pythagoras should work out nicely here :smile:
Reply 2
Original post by L1000
Hey, struggling on vectors, any help with this question will be appreciated very much!

I can do question a, but stuck on part b?

At time t=0, two ice skaters John(J) and Norma(n) have position vectors 40j and 20i metres relative to an origin O at the centre of an ice rink, where i and j are unit vectors perpendicular to each other. John has constant velocity 5im/s and Norma has constant velocity (3i +4j)m/s

a) show that the skaters will collide, and find the time at which the collision takes place.

b) on another occasion, John has position vector 40j metres. He wishes to skate in a straight line to the point with position vector 30i metres. Given that his speed is contant at 5m/s, find his velocity.

I know the equation r = r(position vector) + vt

So I sub in the numbers: 30i = 40j + vt?

I'm not sure on where to go from there, I'm thinking something with pythag since its speed, but sort of confused...

Thanks in advance


Rearrange to get vt=30i40j vt = 30i - 40j .

Now we can take the magnitude of both sides vt=30i40j|vt| = |30i - 40j|. This gives 5t=30i40j 5t = |30i - 40j| from which you find the value of tt.

Finally, dividing the original equation by tt on both sides gives you the velocity.

Note that you can probably do this problem by inspection. It just involves a 3-4-5 triangle.
Reply 3
Original post by davros
You need a vector from 40j to 30i with magnitude 5. Looks like Pythagoras should work out nicely here :smile:



Original post by Brister
Rearrange to get vt=30i40j vt = 30i - 40j .

Now we can take the magnitude of both sides vt=30i40j|vt| = |30i - 40j|. This gives 5t=30i40j 5t = |30i - 40j| from which you find the value of tt.

Finally, dividing the original equation by tt on both sides gives you the velocity.

Note that you can probably do this problem by inspection. It just involves a 3-4-5 triangle.


Thank you I've got the answer now :smile:

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