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M1 Vectors

Hey, I'm wondering if someone could give me some help with a question on vectors from AS Mechanics, M1.

A particle P moves with constant accceleration (2i-3j)ms-2. At time t seconds, it's velocity is
v ms-1. When t = 0, v = -2i+7j.

a) Find the value of t when P is moving parallel to the vector i.
b) Find the speed of P when t = 3.
c) Find the angle between the vector j and the direction of motion of P when t = 3.

I get that when P is parallel to i, P's j value = 0? I think?

Thank you in advance :smile:

(edited 9 years ago)

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Reply 1

chemlover12

Original post by AJC98

To work out velocity from acceleration, you integrate. But then you will have an extra C term, When t = 0, v = -2i+7j, so they gave you this to work that out. Have you covered integration yet in your core maths class?

Reply 2

TLHroolz

Original post by AJC98

a) There will only be a component of i velocity, so yes, the j component would be 0. Use v=u+at

b) again, use v=u+at and plug t=3 in, then take the magnitude

c) use i and j components from b, but remember its from the j vector that the angle is coming from

Original post by chemlover12

Using calculus in M1 is unnecessary. Not needed until M2

Posted from TSR Mobile

(edited 9 years ago)

Reply 3

AJC98

Original post by TLHroolz

Ah right ok, so v=u+at rearranges to t = (v+u)/a. u=-2i+7j, a = 2i-3j and t=0? what's v?

(edited 9 years ago)

Reply 4

TLHroolz

Original post by AJC98

Ah right ok, so v=u+at rearranges to t = (v+u)/a. u=-2i+7j, a = 2i-3j and t=0? what's v?

It'll be 0j plus anything i
And v-u not v+u
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Reply 5

AJC98

Original post by TLHroolz

It'll be 0j plus anything i
And v-u not v+u
Posted from TSR Mobile

So substituting the values into the formula you'd end up with
t=(v-u)/a

0=(v-(-2i+7j))/2i-3j

0=(v+2i+7j)/2i-3j

Now what? It may seem like I'm getting you to do my homework but I genuinely don't understand, sorry to be a pain

(edited 9 years ago)

Reply 6

davros

Original post by AJC98

You can't divide vectors! And why have you put t = 0 - you're supposed to be finding t!

Start from v = u + at (or v = u + ta which is equivalent and is a slightly more usual way of writing things when a is a vector and t is a scalar) and plug in the values you're given for u and a.

Let us know what equation you get at this point.

Reply 7

AJC98

Original post by davros

Right I got:
v=u+ta
v=(-2i+7j)+t(2i-3j)

Reply 8

davros

Original post by AJC98

Right I got:
v=u+ta
v=(-2i+7j)+t(2i-3j)

Good.

Now the first part of the question wants you to find the value of t when motion is parallel to vector i.

If the vector v is parallel to vector i, this means that its j-component must be zero. So if you rewrite your equation for v as

v = (something)i + (something else)j

you need the "something else" to be 0. This should allow you to find t.

Reply 9

AJC98

Original post by davros

How would you rewrite the equation in that way? When you expand those brackets you end up with

v = -2i+7j +2ti + 3tj

But what do I do now?

Reply 10

davros

Original post by AJC98

How would you rewrite the equation in that way? When you expand those brackets you end up with

v = -2i+7j +2ti + 3tj

But what do I do now?

You can just collect the i- and j- terms together, so the i- component is:

(-2 + 2t)i

(remember that t is just a scalar (number) so -2 + 2t is also just some number)

Do the same for the j-component and then follow my previous advice :smile:

EDIT: I think you meant to write -3tj in your first line :smile:

Reply 11

AJC98

Original post by davros

EDIT: I think you meant to write -3tj in your first line :smile:

So you end up with i(-2+2t) and j(7-3t)

v = (-2+2t)i + (7-3t)j

I can only imagine how painful this must be for you, but what now? I'm still quite lost... the (7-3t) has to equal 0??

Reply 12

davros

Original post by AJC98

So you end up with i(-2+2t) and j(7-3t)

v = (-2+2t)i + (7-3t)j

I can only imagine how painful this must be for you, but what now? I'm still quite lost... the (7-3t) has to equal 0??

That's right! So if 7 - 3t = 0 what is the value of t?

Reply 13

AJC98

Original post by davros

That's right! So if 7 - 3t = 0 what is the value of t?

7-3t = 0
7 = 3t
t = 7/3

but why do you ignore the rest of the equation? the "(-2+2t)i" part?

Reply 14

davros

Original post by AJC98

7-3t = 0
7 = 3t
t = 7/3

but why do you ignore the rest of the equation? the "(-2+2t)i" part?

Because it doesn't matter what it is!

You're told that the motion is parallel to the i-direction. That means that the velocity vector looks like v = ki for some scalar k, but you don't actually care what k is (unless they specifically ask you to find its value at the time in question).

Reply 15

AJC98

Original post by davros

Right!

So if the motion was parallel to the j direction, you'd find t for -2+2t = 0?

Reply 16

davros

Original post by AJC98

Right!

So if the motion was parallel to the j direction, you'd find t for -2+2t = 0?

That's correct :smile:

Reply 17

AJC98

Original post by davros

That's correct :smile:

Nice one

thanks!!

Reply 18

AJC98

Ok so now I'm stuck on c) :frown:

for b) My working was:

v = u+at
v = (-2i+7j)+3t(2i-3j)
v = -2i+7j+6it-9it
v = (-2+6t)i + (7-9t)j
The magnitude of a vector is the speed, so
Speed = sq root of (-2+6t)squared+(7-9t)squared
Speed = -2+6t+7-9t
Speed = -3t+5

I don't know if this is right, there's some guesswork in there...

And for c) I've said that

Vector j = 7-9t
Direction of Motion = -2+6t

And now I don't know what to do... I know I'll be using trigonometry somewhere but I don't know how to go about doing that

Thanks :smile:

Reply 19

davros