Motion In One Dimension Question 4
Question 4 - 2024 (27 Jan Shift 1)
A particle starts from origin at $\mathrm{t}=0$ with a velocity $5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ and moves in $\mathrm{x}-\mathrm{y}$ plane under action of a force which produces a constant acceleration of $(3 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}^{2}$. If the $x$-coordinate of the particle at that instant is $84 \mathrm{~m}$, then the speed of the particle at this time is $\sqrt{\alpha} \mathrm{m} / \mathrm{s}$. The value of $\alpha$ is
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Answer: (673)
Solution:
$\mathrm{u}{\mathrm{x}}=5 \mathrm{~m} / \mathrm{s} \quad \mathrm{a}{\mathrm{x}}=3 \mathrm{~m} / \mathrm{s}^{2} h \mathrm{x}=84 \mathrm{~m}$
$\mathrm{v}{\mathrm{x}}^{2}-\mathrm{u}{\mathrm{x}}^{2}=2 \mathrm{ax}$
$\mathrm{v}_{\mathrm{x}}^{2}-25=2(3)(84)$
$\mathrm{V}_{\mathrm{x}}=23 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}{\mathrm{x}}-\mathrm{u}{\mathrm{x}}=\mathrm{a}_{\mathrm{x}} \mathrm{t}$
$\mathrm{t}=\frac{23-5}{3}=6 \mathrm{~s}$
$\mathrm{v}{\mathrm{y}}=0+\mathrm{a}{\mathrm{y}} \mathrm{t}=0+2 \times(6)=12 \mathrm{~m} / \mathrm{s}$
$\mathrm{v}^{2}=\mathrm{v}{\mathrm{x}}^{2}+\mathrm{v}{\mathrm{y}}^{2}=23^{2}+12^{2}=673$
$\mathrm{v}=\sqrt{673} \mathrm{~m} / \mathrm{s}$
