A particle starts from the origin at t=0 with | vedclass

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(A) Given: Initial velocity $\vec{u} = 5 \hat{i} 	ext{ m/s}$,acceleration $\vec{a} = 3 \hat{i} + 2 \hat{j} 	ext{ m/s}^2$,and displacement $x = 84

Vedclass · Accepted Answer

$673$

Vedclass · Answer

(A) Given: Initial velocity $\vec{u} = 5 \hat{i} 	ext{ m/s}$,acceleration $\vec{a} = 3 \hat{i} + 2 \hat{j} 	ext{ m/s}^2$,and displacement $x = 84 	ext{ m}$.First,consider the motion along the $x$-axis: $u_x = 5 	ext{ m/s}$,$a_x = 3 	ext{ m/s}^2$,$x = 84 	ext{ m}$.Using the equation $v_x^2 - u_x^2 = 2 a_x x$:$v_x^2 - 5^2 = 2(3)(84)$$v_x^2 - 25 = 504$$v_x^2 = 529 \implies v_x = 23 	ext{ m/s}$.Next,find the time $t$ using $v_x = u_x + a_x t$:$23 = 5 + 3t \implies 3t = 18 \implies t = 6 	ext{ s}$.Now,consider the motion along the $y$-axis: $u_y = 0$,$a_y = 2 	ext{ m/s}^2$,$t = 6 	ext{ s}$.$v_y = u_y + a_y t = 0 + 2(6) = 12 	ext{ m/s}$.The speed $v$ is given by $\sqrt{v_x^2 + v_y^2}$:$v = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} 	ext{ m/s}$.Comparing this with $\sqrt{\alpha}$,we get $\alpha = 673$.

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