Mechanics

Daniel123 · 2007-09-18 02:51:18

I don't know how to do this. Rather than doing it for me, can someone please explain to me how to do it, so I can do it myself?

A particle has position vector 2i + j initially and is moving with a speed of 10ms-¹ in the direction 3i - 4j. Find its position vector when t=3 and the distance it has travelled in those 3 seconds.

Thanks!!

mathsyperson · 2007-09-18 03:43:12

First work out the velocity vector.
You've got a vector indicating the direction, but it's got the wrong magnitude. Work out what its magnitude is (with Pythagoras) and then scale it so that the magnitude becomes 10.

Once you've got that you'll know how much the particle moves in 1 second, so just multiply the velocity vector by 3 and add that to the initial position.

The second part of the question can be answered without using vectors. You're given the speed and time in the question, so the distance is simple to work out.

Daniel123 · 2007-09-18 04:31:47

I started doing what you said, but I get totally ridiculous numbers, that can't possibly lead to the answer in the back of my book. What am I doing wrong?

NOTE: The bottom co-ordinate is meant to say (3,-4)

Last edited by Daniel123 (2007-09-18 04:32:35)

mathsyperson · 2007-09-18 04:35:50

Velocity vectors are measured from the origin, not from the initial point. Forget about 2i+j until you've found the velocity in terms of i and j.

luca-deltodesco · 2007-09-18 04:50:21

: A particle has position vector 2i + j initially and is moving with a speed of 10ms-¹ in the direction 3i - 4j. Find its position vector when t=3 and the distance it has travelled in those 3 seconds. :

|3i - 4j| = 5, so the velocity is: 6i - 8j

no acceleration:

s[3] = [2i+j]+3[6i-8j] = 20i - 23j

distance travelled = 3*10 = 30m

Last edited by luca-deltodesco (2007-09-18 04:51:00)

JaneFairfax · 2007-09-18 05:42:49

Lets do it this way. The speed is 10 ms[sup]−1[/sup]; ∴ (assuming no acceleration) the particle travels 30 m in 3 s.

You want to find a vector with magnitude 30 and parallel to 3i−4j. To do this, you find a unit vector parallel to 3i−4j and multiply it by 30. A unit vector parallel to 3i−4j is

So the vector we want is 30×(3i−4j)∕5 = 6×(3i−4j) = 18i−24j. Now, add that to the particles initial position: 18i−24j + 2i+j = 20i−23j. Thats its position vector after 3 seconds.

Last edited by JaneFairfax (2007-09-18 05:49:01)

Daniel123 · 2007-09-18 06:36:17

The bit I don't get is why you add it to the particle's initial position, and why the direction comes from the origin? :s

luca-deltodesco · 2007-09-18 06:45:30

direction doesn't come from the origin, it doesn't come from anywhere, its just a direction.

you add it to the initial position, because with the velocity, you calculate the displacement, the displacement is how far the particle has been displaced - from its initial position, so its final position is its displacement + its intial position

Daniel123 · 2007-09-18 07:22:42

Ahh.

Thanks all.

Daniel123 · 2007-09-18 09:33:17

Ok I have a few problems in here.. sorry its long and wordy.

(In this question the unit vectors i and j are due east and due north respectively)
At noon a ship S is 600m due north of an observation point O and a speedboat B is 120m due north of the same point. The ship S has a constant velocity (7i + 8j)ms-¹ and the speedboat B has a constant velocity of (7i + 24j)ms-¹

a) Write down the position vectors of S and B at time t seconds after noon.

S: 600j + (7i +8j)t
B: 120j + (7i+24j)t

b) Show that S and B will collide and find the time when this collision occurs and the position vector of the point of collision.

Show they will collide: For them to collide, poisition vectors must be equal, therefore:

600j + (7i +8j)t = 120j + (7i+24j)t
600j + 7it + 8jt = 120j + 7it + 24jt
i(7t) + j(8t + 600) = i(7t) + j(24t + 120)
i's are equal.. what do i do here about the j's?

Find the time:
8t + 600 = 24t + 120
16t = 480
t = 30 seconds

Find position vector:
600j + (7i + 8j)30 = (210i + 840j)m

c) In order to prevent a collision, 15s after noon S changes its velocity to (7i + 30j)ms-¹. Find the distance between S and B 30s after noon.

S: 600j + (7i +8j)t = 600j + (7i + 8j)30 = 210i + 840j
B: 120j + (7i+24j)15 + (7i + 30j)15 = 210i + 1170j

Distance between S and B is therefore 330j... but this is wrong. Why?

A quick response would be appreciated!
Thanks!

John E. Franklin · 2007-09-18 10:02:18

20 and -23 is right because 3-4-5 triangle has hypotenuse of 5, so 10m/s, it goes
two fives, and times 3 seconds is six of the 3-4-5 triangles.
So 6x3 is 18 and 6x(-4) is -24. Add 18 to 2, and subtract 24 from 1 for -23.
Put new question in new post?

JaneFairfax · 2007-09-18 11:55:08

Daniel123 wrote:

c) In order to prevent a collision, 15s after noon S changes its velocity to (7i + 30j)ms-¹. Find the distance between S and B 30s after noon.
S: 600j + (7i +8j)t = 600j + (7i + 8j)30 = 210i + 840j
B: 120j + (7i+24j)15 + (7i + 30j)15 = 210i + 1170j
Distance between S and B is therefore 330j... but this is wrong. Why?

Its wrong because S is the boat that changes direction, not B.

Daniel123 · 2007-09-18 22:34:03

Oops.. but it still works out to 330j.... but the answer in my book is 90m. :s

mathsyperson · 2007-09-18 23:18:37

I think you've just put a typo in your calculatarr or something. In your working, the line should read:

B: 120j + (7i+24j)15 + (7i + 30j)15 = 210i + 930j

That then gives a diffarrence of 90m.

JaneFairfax · 2007-09-18 23:25:37

In that case, its B that changes velocity, not S.

And as Mathsy pointed out you misculated Bs new position vector.

Last edited by JaneFairfax (2007-09-18 23:26:40)

Math Is Fun Forum

#1 2007-09-18 02:51:18

Mechanics

#2 2007-09-18 03:43:12

Re: Mechanics

#3 2007-09-18 04:31:47

Re: Mechanics

#4 2007-09-18 04:35:50

Re: Mechanics

#5 2007-09-18 04:50:21

Re: Mechanics

#6 2007-09-18 05:42:49

Re: Mechanics

#7 2007-09-18 06:36:17

Re: Mechanics

#8 2007-09-18 06:45:30

Re: Mechanics

#9 2007-09-18 07:22:42

Re: Mechanics

#10 2007-09-18 09:33:17

Re: Mechanics

#11 2007-09-18 10:02:18

Re: Mechanics

#12 2007-09-18 11:55:08

Re: Mechanics

#13 2007-09-18 22:34:03

Re: Mechanics

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Re: Mechanics

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