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(D) Given, mass of the object, $m = 2 \ kg$.Velocity, $v = (8t \hat{i} + 3t^2 \hat{j}) \ m/s$.The acceleration of the object is given by:$a = \frac{dv

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$	an^{-1}(3/4)$

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(D) Given, mass of the object, $m = 2 \ kg$.Velocity, $v = (8t \hat{i} + 3t^2 \hat{j}) \ m/s$.The acceleration of the object is given by:$a = \frac{dv}{dt} = \frac{d}{dt}(8t \hat{i} + 3t^2 \hat{j}) = (8 \hat{i} + 6t \hat{j}) \ m/s^2$.According to Newton's second law of motion, the net force on the object is:$F = ma = 2(8 \hat{i} + 6t \hat{j}) = (16 \hat{i} + 12t \hat{j}) \ N$.The magnitude of the force is given as $|F| = 20 \ N$.$|F| = \sqrt{16^2 + (12t)^2} = 20$.Squaring both sides:$256 + 144t^2 = 400$.$144t^2 = 144 \implies t^2 = 1 \implies t = 1 \ s$.At $t = 1 \ s$, the force vector is:$F = 16 \hat{i} + 12(1) \hat{j} = (16 \hat{i} + 12 \hat{j}) \ N$.The angle $	heta$ made by the force vector with the positive $X$-axis is:$	an 	heta = \frac{F_y}{F_x} = \frac{12}{16} = \frac{3}{4}$.$	heta = 	an^{-1}\left(\frac{3}{4}ight)$.

The velocity of an object of mass $2 \ kg$ is given by $v = (8t \hat{i} + 3t^2 \hat{j}) \ m/s$, where $t$ is time in seconds. What will be the direction of the net force on the object relative to the positive direction of the $X$-axis, at the instant when its magnitude is $20 \ N$?

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