Question

a particle moves with a speed of 10 ms from the
point (2,-2) in the direction 3i+4j. The position
vector after 3 s is
(1) 181 +24;
(2) 201 + 22;
(3) 101 - 117
(4) 201 – 247​

Answer 1

Hii Dear,

● Answer -

(2) 20i + 22j

● Explaination -

Unit vector in direction of 3i + 4j is given by -

Unit vector = 3i+4j / √(3²+4²)

Unit vector = 3i+4j / √25

Unit vector = 3i/5 + 4j/5

Magnitude of displacement in 3 s is calculated as -

Displacement(magnitude) = velocity (magnitude) × time

Displacement(magnitude) = 10 × 3

Displacement(magnitude) = 30 m

Therefore, displacement in 3 s will be -

Displacement = magnitude × unit vector

Displacement = 30 × (3i/5 + 4j/5)

Displacement = 18i + 24j

But, we know that displacement is calculated by -

Displacement = Final position - Initial position

18i + 24j = Final position - (2i - 2j)

Final position = 18i + 24j + 2i - 2j

Final position = 20i + 22j

Therefore, final position of the particle will be 20i + 22j.

Thanks dear...