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Mechanics 1 Help Please - Position vectors

How would I go about answering this question I have no idea how to answer this.

A particle moving with speed vms-1 in direction d has velocity vector v . Find v when v=10 and d= 3i-4j
Reply 1
need more detail please becuase with this information u only have u and s value does it say anything about dropped from height above ground??? if so we now have a and can use v^2=u^2+2as sorry hope that helps please give full question.
Reply 2
The velocity vector v has the same direction as the direction vector d.

i.e. d = 3i - 4j and v = si + tj, where s and t are in the same ratio as 3 and -4. i.e. s/t = - 3/4.

The speed of v is given by v = sqrt(s^2 + t^2).
Reply 3
Original post by steve10
The velocity vector v has the same direction as the direction vector d.

i.e. d = 3i - 4j and v = si + tj, where s and t are in the same ratio as 3 and -4. i.e. s/t = - 3/4.

The speed of v is given by v = sqrt(s^2 + t^2).


So how would I change that speed into a vector ?
Original post by Olive123
How would I go about answering this question I have no idea how to answer this.

A particle moving with speed vms-1 in direction d has velocity vector v . Find v when v=10 and d= 3i-4j


What is the magnitude of the direction? It's (3^2 + 4^2)^1/2 right? So 5m. But you know it travels 10 metres per second, and so what vector does it travel every second?
Reply 5
Original post by hassi94
What is the magnitude of the direction? It's (3^2 + 4^2)^1/2 right? So 5m. But you know it travels 10 metres per second, and so what vector does it travel every second?


Does it travel that direction vector every second ?
Original post by Olive123
Does it travel that direction vector every second ?


Well no, almost. The direction vector has magnitude 5m. But it travels 10 metres every second. (it must travel at a multiple of the direction vector, every second, but what is that multiple?)
Reply 7
Original post by Olive123
So how would I change that speed into a vector ?


set the speed, to the value given, in the equation I gave, and also use the ratio I gave to solve for s or t.
Reply 8
Original post by hassi94
What is the magnitude of the direction? It's (3^2 + 4^2)^1/2 right? So 5m. But you know it travels 10 metres per second, and so what vector does it travel every second?


Actually as it is 5m it travels two direction vectors evey second ?

But how does that then equal a velocity vector :s soo confusssed :frown:
Original post by Olive123
Actually as it is 5m it travels two direction vectors evey second ?

But how does that then equal a velocity vector :s soo confusssed :frown:


Yep that's right.

So it travels 2*(3i-4j) each second AKA (6i - 8j)ms^-1

Make sense?
Reply 10
The velocity vector is v = si + tj. Solve for s and t.
The velocity vector is made up of a unit vector multiplied by a magnitude (the speed). The unit vector in the direction of the velocity is given by d divided by its magnitude.
Reply 12
Original post by hassi94
Yep that's right.

So it travels 2*(3i-4j) each second AKA (6i - 8j)ms^-1

Make sense?


Yes sort of, I just need to practice I guess to get my head around it. Thanks
Original post by Olive123
Yes sort of, I just need to practice I guess to get my head around it. Thanks


The below poster explains it without taking 'shortcuts' as I did, if that helps :smile:

Original post by notastampcollector
The velocity vector is made up of a unit vector multiplied by a magnitude (the speed). The unit vector in the direction of the velocity is given by d divided by its magnitude.
Reply 14
Thank you everyone I really appreciate your help !!
Original post by Olive123
How would I go about answering this question I have no idea how to answer this.
A particle moving with speed vms-1 in direction d has velocity vector v . Find v when v=10 and d= 3i-4j

Velocity =(speed/magnitude of direction vector)×direction vector

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