A particle moves in the x-y plane with velocity {V} = {xi}+{ytj} ; At t = {x{3}}{y} , the magnitudes of tangential and normal accelerations, respectively are

Correct Answer is : 3y2,y2{{3}y}{2},y/223y,2y

Correct Answer is : 3y2,y2{{3}y}{2},y/223y,2y

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AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Section: Physics

Question : 83 of 160

Marks: +1, -0

\vec{V}

\widehat{xi}+\widehat{ytj}

\frac{x\sqrt{3}}{y}

Solution:

Given, the expression of the velocity is,

v=x\hat{i}+yt\hat{j}

The magnitude of the velocity is,

|v|=\sqrt{x^{2}+y^{2}t^{2}}

The magnitude of the tangential acceleration is,

a_{t}=\frac{d|v|}{dt}

a_{t}=\frac{d\sqrt{x^{2}+y^{2}t^{2}}}{dt}

a_{t}=\frac{ty^{2}}{\sqrt{x^{2}+y^{2}t^{2}}}

The tangential acceleration at

t=\frac{x\sqrt{3}}{y}

is,

a_{t}=\frac{\left(\frac{\sqrt{3}}{y}\right)y^{2}}{\sqrt{x^{2}+y^{2}\left(\frac{x\sqrt{3}}{y}\right)^{2}}}

=\frac{\sqrt{3}y}{2}

The total acceleration is,

a=\frac{dv}{dt}

a=\frac{d}{dt}(x\hat{i}+yt\hat{j})

a=y\hat{j}

The magnitude of the normal acceleration is,

|a_{n}|=\sqrt{a^{2}-a_{b}^{2}}

=\sqrt{y^{2}-\left(\frac{y^{4}t^{2}}{x^{2}+y^{2}t^{2}}\right)}

=\sqrt{\frac{x^{2}y^{2}}{x^{2}+y^{2}t^{2}}}

=\frac{xy}{\sqrt{x^{2}+y^{2}t^{2}}}

The magnitude of normal acceleration at

t=\frac{x\sqrt{3}}{y}

is,

|a_{n}|=\frac{xy}{\sqrt{(x^{2}+y^{2})\frac{3x^{2}}{y^{2}}}}

=\frac{y}{2}

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