AS Vector Question

In this question i and j are horizontal unit vectors and at right angles to each other.
At 12:00 a helicopter A sets out from its base 0 and flies with speed 120 km h-1 in the direction of the vector 3i+4j.

a) Find the velocity vector of A
At 12:20 that day another helicopter B sets out from O and flies with speed 150 km h -1in 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-1 in a straight line to B.
e) Find the position vector of B from A at 13:30
f) Determine the time when A reaches B.

If anyone is bored some help would be much appreciated because I was never taught any of this :frown:

Reply 1

D-Day

I'll help, I'm just going to take a moment to compose my reply to all parts at once.

Reply 2

D-Day

Drawing pictures helps immensely.

a) Says to find the velocity of the helicopter. Velocity is composed of a magnitude (speed) and a direction (angle). You know the speed because it's given to you. Using the fact that this is a 3-4-5 right triangle, you can find the magnitudes of the x and y components of the 120 km/h speed. Then use one of the three trig functions (sin, cos or tan) to find the angle.

b) Same as a). From the Pythagorean Theorem we know that the relationship of the sides of this triangle are 7-24-25. Use this to find the magnitudes of x and y and then use a trig function to find the angle.

c) Since the two helicopters fly in a straight line, the angles here will be unchanged from parts a) and b). Use the fact that distance equals velocity (speed) times time to find the magnitude of the position vectors after one hour.

d) For this it will help to overlay your two pictures onto one picture. You have the 3-4-5 triangle with the angle between the vector of helicopter A known (from part a)) and you have the 7-24-25 triangle from b) also with the angle known. Do you see here that if you subtract the angle for B from the angle for A you get the angle between them? Then use the law of cosines to find the length of the side opposite this angle. This is the distance you need.

e) The position vector from A to B is just the distance between them @ some angle. If you think about the triangle that's formed with the point O (where they both set off), the point that A is at, and the point B is at, as the three vertices, this is fairly easy. You can find the length of the side from the point O to the point of B by using your 7-24-25 triangle again and you can use d=vt again to find the length of the side from point O to point A at time 1.5 hours. You know the angle between these two sides by adding the angle from part a of this problem to the angle from part b of this problem. Use the law of cosines to find the length of the side between A and B. Then use the law of sines to find the angle between the bottom side of the triangle and the position vector of A to B. This angle will actually be negative because the position vector is pointing down (below horizontal).

f) You know the distance between A and B and you know the speed at which A is moving toward B (given to you). d=vt again to find time.

Edit: I am quite certain that you are going to have many, many questions about my explanation. Feel free to post them here, or PM me :smile:

Reply 3

silent ninja

D-Day

f) I'm not sure what this one means, unless you're missing something from the problem. If A and B are flying in a straight line at some angle away from each other (as in our pictures), they never meet.