A particle moves over a xy plane with a const | vedclass

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(A) Given: Acceleration $\vec{a} = 4 \hat{i} + 4 \hat{j} \, m \, s^{-2}$, Initial velocity $\vec{u} = 4 \hat{i} \, m \, s^{-1}$.Displacement along the

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$4 \sqrt{5} \, m \, s^{-1}$

Vedclass · Answer

(A) Given: Acceleration $\vec{a} = 4 \hat{i} + 4 \hat{j} \, m \, s^{-2}$, Initial velocity $\vec{u} = 4 \hat{i} \, m \, s^{-1}$.Displacement along the $x$-axis is $s_x = 6 \, m$.Using the equation of motion $s_x = u_x t + \frac{1}{2} a_x t^2$:$6 = 4t + \frac{1}{2}(4)t^2$$6 = 4t + 2t^2 \Rightarrow 2t^2 + 4t - 6 = 0 \Rightarrow t^2 + 2t - 3 = 0$.Solving for $t$: $(t+3)(t-1) = 0$. Since $t > 0$, we have $t = 1 \, s$.Now, find the velocity components at $t = 1 \, s$:$v_x = u_x + a_x t = 4 + 4(1) = 8 \, m \, s^{-1}$.$v_y = u_y + a_y t = 0 + 4(1) = 4 \, m \, s^{-1}$.The resultant speed $v = \sqrt{v_x^2 + v_y^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4 \sqrt{5} \, m \, s^{-1}$.

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