HARD A LEVEL motion in 2D question

The unit vectors i and j are oriented due east and due north, respectively.
At time t = 0 s , a particle of mass 4 kg , is sighted at the point A with position vector (−17i − 50j) m and moving with constant velocity (−2i + 2j) ms−1 .
At time t =10 s , two constant forces F1 and F2 each of magnitude 50 N , begin to
act on the particle until it passes through the point B , after a further period of 10 s . It is further given that
• F1 is acting in the direction (3i − 4j) .
• F2 is acting in the direction (−7i + 24j).
Determine, correct to the nearest m , the distance between A and B .

the answer is 213m.
can someone explain how to get this answer please ?

Steps:

- Figure out where the particle is after the first 10 seconds as it moves with const. velocity -2i + 2j

- Now two forces are acting on the particle. So there is a resultant force for you to find. Hence use $F= ma$ to determine the acceleration vector.

- Use SUVAT to determine the where the particle ends up after the next 10 seconds. This is your point B.

- Now just find the distance between the two points. We know A from the context.

For some reason the markscheme mulitiplies the direction of F2 by 2 and not 10, why is that?

F2 is a 7:24:25 triangle so 14:48:50 where the 50 is the hypotenuse or force magnitude. F1 is 3:4:5 or 30:40:50.

oh my gosh THANK YOU SO MUCH, that made more sense. So it's a scale factor of the magnitute and it isn't being multiplied by time?