(C) Given: Initial velocity $\vec{u} = 3.0 \hat{i} \; m/s$,acceleration $\vec{a} = (6.0 \hat{i} + 4.0 \hat{j}) \; m/s^2$,and initial position $(0, 0)$

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(C) Given: Initial velocity $\vec{u} = 3.0 \hat{i} \; m/s$,acceleration $\vec{a} = (6.0 \hat{i} + 4.0 \hat{j}) \; m/s^2$,and initial position $(0, 0)$ at $t=0$.For motion in the $y$-direction:$y = u_y t + \frac{1}{2} a_y t^2$$32 = 0 \times t + \frac{1}{2} (4.0) t^2$$32 = 2 t^2$$t^2 = 16 \implies t = 4 \; s$.For motion in the $x$-direction:$x = u_x t + \frac{1}{2} a_x t^2$$x = (3.0)(4) + \frac{1}{2} (6.0)(4)^2$$x = 12 + 3.0 \times 16$$x = 12 + 48 = 60 \; m$.Thus,$D = 60$.

Class 11 Physics 2 Motion in a Plane Mix Examples-Motion in Plane

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$A$ 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

JEE MAIN 2020 All JEE MAIN

A
$50$
B
$32$
C
$60$
D
$40$

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