Question

A particle whose speed is 576 m/s moves along the line fromA(1,0,3) to B(3,2,-1).find its
velocity vector in the form of ai+bj + ck
(A) 5i+5– 10 (B) 5î + 10– 10K (C) 5* +59 +10
(D) None of these​

Answer 1

Answer:

Position vector along AB

= 2i+2j-4k

unit vector along AB = 2i+2j-4k/√24

multiply by 5√6

5√6 × (2i+2j-4k)/2√6

= 5i+5j-10

(A)

Answer 2

Explanation:

Displacement will be final position-initial position

9

i

^

−2

i

^

+25

j

^

cap−jcap

=7

i

^

+24

j

cap

Speed =distance /time

$$50=25/time(use distance formula which u studied in 10th)

Time=

2

1

sec

Velocity =distance /time

V=

2

1

2

i

+24

j

cap

V=2(2

i

+24

j

)

Hope it helps you! plz mark it as BRAINLIEST!!