Question

the position vector of the particle is (2i+3j) at t=0. velocity of the particle becomes (3i+6j) m/s at t=2sec. then find the position vector of the particle at t=2sec. what is the x and y cordinate ​

Answer 1

answer is 8i +15j

hope this may help you..

Answer 2

position vector of particle after t = 2 sec is 8i + 15j

The position before of the particle, x_0 = 2i + 3j , at t = 0.

velocity of particle after t = 2sec is v = 3i + 6j

now using formula, x = x_0 + vt

where x is the position vector of particle after t = 2sec.

x = (2i + 3j) + (3i + 6j) × 2

= (2i + 3j) + (6i + 12j)

= 8i + 15j

hence position vector of particle after t = 2 sec is 8i + 15j

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