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Question

# At time t=0 s, a particle starts from rest at position (4 m,8 m). It starts moving towards the positive x-axis with a constant acceleration of 2 m/s2. After 2 s, it undergoes an additional acceleration of 4 m/s2 in the positive y-direction. Find the coordinates of the particle (in m) after the next 5 s.

A
(58,6)
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B
(53,58)
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C
(53,58)
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D
(58,6)
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Solution

## The correct option is B (53,58)Given, Initial coordinates of particle (x0,y0)=(4,8) Initial velocity of the particle (ux,uy)=(0,0) From the data given in the question, Particle is moving along x- axis with a constant acceleration ax=2 m/s2 Therefore, we can use the kinematic formulae. Using v=u+at, we can write that, vx=ux+axt At t=2 sec , vx=0+2×2⇒vx=4 m/s vy=0+0×2⇒vy=0 m/s Position of the particle at t=2 s is given by x1=x0+uxt+12axt2 =4+0+12×(2)×22=8 m y1=y0+uyt+12ayt2 =8+0+12×(0)×22=8 m ⇒ Position of the particle at t=2 s is (x1,y1)=(8,8) m Now, after next 5 seconds, Position of the particle is given by x2=x1+vxt+12axt2 y2=y1+vyt+12ayt2 From the given data , x2=8+(4)×5+12×2×(5)2=53 m y2=8+(0)×5+12×4×(5)2=58 m Hence, the position of the particle is given by (x2,y2)=(53 m,58 m) Thus, option (b) is the correct answer.

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