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Oh God... i think i'm failing Maths... Help me

Decidied to open up a thread dedicated to all my maths problems, cos it looks like the're gonna be plenty....

Right... so the first problem is on M1

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, hohn has position vector 4oj 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.

Please help. It seems like its menat to have areally simplem solution, but it's just not clicking right now....

Reply 1

Mohit_C

Rite, for part a), they both have different position vectors, so you have to find where they will be after t seconds. Distance = speed * time. Therefore for John his distance travelled after t seconds is:
(5i) * t = 5ti

For Norma it's: (3i+4j)t= 3ti+4tj

But don't forget that they both already started at an initial position vector therefore for Norma:
after t seconds: (20+3t)i+(4t)j

For John:
after t seconds: (5t)i+(40t)j.

To prove that they will collide at a particular t, bot i and j vectors will coincide therefore making one equal to the other:
for i's: 5t=20+3t
Therefore: 2t=20, t=10

for j's: 40=4t, t=10
Therefore as both i's and j's vector coincide at t=10, they will collide at t=10.

Reply 2

insparato

Position Vector at time, t = initial position vector + displacement(ie velocity x time).

Reply 3

Babyangel452

aaahhh... right... makes sense...

Reply 4

Babyangel452

In this question I and j are horizontal vectors and at right angles to each other.

At 12.00 a helicopter a sets out from its base O and flies with speed 120 km/h in the direction of the vector 3i + 4 J.

a) find the velocity of A

At 12.20 that day, another helicopter B sets out from O and flies with speed 150 km/h in the direction of the vector 24i+7j.

b) find the velocity vector of B
c) Find the position vectors of A and B at 13.00
d) Calculate the distance of A from B at 13.00

At 13.30 B makes an emergency landing. A immediately changes direction and flies at 120 km/h in the straight line to B.

e) Find the position Vector of B from a at 13.30
f) Determine the time when A reaches B.

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the position vector r of a particle P at time t is given by r= t^2i + (12-t)j. find the value of t when:
a) r is parallel to vecotr i
b) r is parallel to vector i+j

I think my problem with these questions is that i just ain't sure of the rules... i know that if the vectors are parallel, then they are in ratios to each other, were one needs to be multiplied by a number to get the other value....

other than that, i know jack... seriously... please help!

Reply 5

Mohit_C

Well for part a) u have to get 120 km/h into vector form, and you know the direction it's going in. Therefore rt3^2+4^2=5. Something *5 =120
Therefore that something is 24. So multiply the direction's i and j vectors by 24 and it gives you: 72i+96j
The same goes for part b)
In part c) you apply the same principle i used as for your very first question which is what insparato quoted, and instead of t just put the time (hours) after 12 which is 1.
d) the distance is when u put t=1 into the position vector of A and B after t hours. Then once you have the two position vectors u subtract ones i vector from the other, square that and you add that to the square of the difference of the j vector. The final answer root is the distance between them.