Question

One particle starts from point A (5m, 8m) with a velocity of (4i + 7j) m/s.At the same time another particle starts from point B (35m, 2m) with a contant velocity of (-6i + uj) m/s.If the two particles,find the value of u.

Answer 1

Answer:

please complete the question

Answer 2

Answer:

the  value of u is 9

Explanation:

given that the position A is(5m,8m)

∵the position vector of A is (5i^ +8j^)
similarly, position vector of B is (35i^ +2j^)

∵ AB vector is  equal to (30i^ - 6j^)       (By Triangle law of vector addition)

then let the time taken for the particles to collide be t ;

then displacement of particle A is (4t i^ +7t j^)m

and displacement of particle B is (-6t i^ + ut j^)m

∵ adding AB vector and displacement of  B particle we get

[(30-6t) i^ +(ut-6) j^ = (4t i^ +7t j^) ] (By Triangle law of vector addition)

(30-6t) i^ =4t i^  (solving you get t= 3)

(ut-6) j^ = 7t j^ (put the value of t and solve)

then the value of u comes 9

I hope it helps :)