Question

A particle starts moving from the point (1,2,3) under velocity 1i^+2j^+3k^) for two seconds. After that it mives under velocity (1i^-2j^+3k^) for next two seconds. what is the coordinate of the final point ?


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Answer 1

Given info : A particle starts moving from the point ( 1, 2, 3) under velocity (1i^ + 2j^ + 3k^) for two seconds. After that it moves under velocity (1i^ - 2j^ + 3k^) for next two seconds.

solution : initial position of particle, s₀ = (1i + 2j + 3k)

initial velocity, u = (1 i + 2j + 3k)

after two second, displacement of particle, s₁ = u.t

= (1 i + 2j + 3k) × 2 = (2i + 4j + 6k)

now 2s ≤ t ≤ 4s, particle moves with v = (1i - 2j + 3k) velocity.

now displacement of particle during 2sec to 4 sec, s₂ = vt

= (1i - 2j + 3k) × 2

= 2i - 4j + 6k

now total displacement of particle = s₀ + s₁ + s₂

= (1i + 2j + 3k) + (2i + 4j + 6k) + 2i - 4j + 6k

= 5i + 2j + 15k

Therefore the final position of particle is (5 , 2 , 15).

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Answer 2

Answer:

5icap +2jcap+12kcap is the right answer