1. ## Position vectors

urgent help, A Level Maths Mechanics

A particle has position vector 2i + j initially and is moving with speed 10ms^-1 in the direction 3i - 4j. Find its position vector when t = 3.

2. Originally Posted by Carl Feltham
urgent help, A Level Maths Mechanics

A particle has position vector 2i + j initially and is moving with speed 10ms^-1 in the direction 3i - 4j. Find its position vector when t = 3.
Unit vector in direction of 3i - 4j is (3i - 4j)/5. Therefore v = 10 (3i - 4j)/5 = 6i - 8j.

Therefore dx/dt = 6i - 8j => x = 6ti - 8tj + C.

When t = 0, x = 2i + j therefore C = 2i + j.

Therefore x = (6t + 2) i + (1 - 8t) j.

3. ## i do not understand

Unit vector in direction of 3i - 4j is (3i - 4j)/5. Therefore v = 10 (3i - 4j)/5 = 6i - 8j.

Therefore dx/dt = 6i - 8j => x = 6ti - 8tj + C.

When t = 0, x = 2i + j therefore C = 2i + j.

Therefore x = (6t + 2) i + (1 - 8t) j.

I understand the rest, bar how you managed to come about the two statements highlighted in red. Why is the first divided by and, and the second doubled?

4. Originally Posted by Carl Feltham
Unit vector in direction of 3i - 4j is (3i - 4j)/5. Therefore v = 10 (3i - 4j)/5 = 6i - 8j.

Therefore dx/dt = 6i - 8j => x = 6ti - 8tj + C.

When t = 0, x = 2i + j therefore C = 2i + j.

Therefore x = (6t + 2) i + (1 - 8t) j.

I understand the rest, bar how you managed to come about the two statements highlighted in red. Why is the first divided by and, and the second doubled?
Remember that velocity is a vector and vectors have a magnitude and a direction. The direction of v is the unit vector in the direction of 3i - 4j. The magnitude of v is its speed 10 m/s. So you multiply the unit vector by 10 (the magnitude of the velocity) to get the velocity vector.

How to get a unit vector: Divide the given vector by its magnitude. The magnitude of 3i - 4j is 5. Therefore you divide 3i - 4j by 5 to get a unit vector in the direction of 3i - 4j.

10 (3i - 4j)/5 = 2 (3i - 4j) = 6i - 8j.