#### 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 $\left ( 6.0\hat{i}+4.0\hat{j} \right )m/s^{2}.$ The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is : Option: 1 60 Option: 2 50 Option: 3 32 Option: 4 40

$\\ \text{Given}:\\ \text{Taking motion along y -axis} \\ \text{Displacement along y-axis}(s_{y})=32 \ m\\ \text{Acceleration along y-axis},(a_{y})=4\ m/s^{2}\\ \text{Initial velocity along y-axis},(u_{y})=0 \ m/s\\\\ \text{From 2nd equation of motion}, \\ s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2} \\ \\ \Rightarrow 32=0(t)+\frac{1}{2}(4)(t^{2})\\ \\ \Rightarrow 32=2t^{2}\\ \\\Rightarrow t=4\ s$

\\ \text{Now taking motion along x- axis.} \\ \text{Displacement along x-axis} \left(\mathrm{s}_{\mathrm{x}}\right)= ? \\ \text{Acceleration along x-axis },(a_{x})=6 \ m/s^{2} \\ \text{initial velocity along x-axis} \left(\mathrm{u}_{\mathrm{x}}\right)=3 \ m/s \\ \begin{aligned} &\text { So, } \mathrm{S}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2} \\ &=3 \times 4+\frac{1}{2} \times 16 \\ &=12+48 \\ &=60 \mathrm{~m} \end{aligned}
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