A particle starts from the origin at t=0 with an initial velocity of 3.0i^m/s and moves in the x-y plane with a constant acceleration 6.0i^+4.0j^m/s^2. The x- coordinate of the particle at the instant when its y- coordinate is 32m is D meters. The value of D is:

Correct answer: (C) 60. S_y=u_yt+1/2a_yt^2 32=0+1/2×4t^2⇒t=4 s S_x=u_xt+1/2a_xt^2 =3×4+1/2×6×16 =60 m

Q69 / 75 · JEE Main 2020 (09 Jan Shift 2) · Physics

A particle starts from the origin at $t = 0$ with an initial velocity of $3.0 \hat{i} m / s$ and moves in the $x - y$ plane with a constant acceleration $(6.0 \hat{i} + 4.0 \hat{j}) m / s^{2} .$ The $x -$ coordinate of the particle at the instant when its $y -$ coordinate is $32 m$ is $D$ meters. The value of $D$ is:

S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2}

32 = 0 + \frac{1}{2} \times 4 t^{2} \Rightarrow t = 4 s

S_{x} = u_{x} t + \frac{1}{2} a_{x} t^{2}

= 3 \times 4 + \frac{1}{2} \times 6 \times 16

= 60 m

Hence, the correct answer is option: 'C'.