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

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Question

$ \mathrm{u}_{\mathrm{x}}=5 \mathrm{~m} / \mathrm{s} \quad \mathrm{a}_{\mathrm{x}}=3 \mathrm{~m} / \mathrm{s}^2 \quad \mathrm{x}=84 \mathrm{~m} $
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Vedclass · Accepted Answer

$673$

Vedclass · Answer

$ \mathrm{u}_{\mathrm{x}}=5 \mathrm{~m} / \mathrm{s} \quad \mathrm{a}_{\mathrm{x}}=3 \mathrm{~m} / \mathrm{s}^2 \quad \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 	imes(6)=12 \mathrm{~m} / \mathrm{s} 4$
$ \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}$

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