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A particle moves in the xy - plane with velocity v = xhati+ythatj. At t = (xsqrt(3))/(y) the magnitudes of tangential and normal accelerations respectively are |
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Answer» `(sqrt(3Y))/(2),(y)/(2)` So`|v|=v=sqrt(x^(2)+y^(2)t^(2))` Magnitude of tangential acceleration is `a_(t)=(dv)/(DT)=(0+2ty^(2))/(2sqrt(x^(2)+y^(2)t^(2)))` or `a_(t)=(ty^(2))/(sqrt(x^(2)+y^(2)t^(2)))` Now substituting `t=(xsqrt(3))/(y)` we get `((xsqrt(3))/(y)xxy^(2))/(sqrt(x^(2)+y^(2)xx((x^(2)xx3)/(y^(2)))))` `implies a_(t)=(sqrt(3y))/(2)` Also total acceleration of particle is a `=(dv)/(dt)` `implies a=(d)/(dt)(xhati+yt hatj)` `implies a=yhatj` or magnitude of total acceleration is `a=|a|=y` Hence normal acceleration is `a_(n)=a_("total")-a_("tangential ")` So `a_(n)=|a_(n)|=` magnitude of normal acceleration `=sqrt(a^(2)-a_(1)^(2))=sqrt(y^(2)-((y^(4)y^(2))/(x^(2)+y^(2)t^(2))))` `=sqrt((x^(2)y^(2))/(x^(2)+y^(2)t^(2)))=(xy)/(sqrt(x^(2)+y^(2)t^(2))` Now with t `=(xsqrt(3))/(y)` we get `a_(n)=|a_(n)|=(xy)/(sqrt(x^(2)+y^(2).(3x^(2))/(y^(2))))=(xy)/(2x)=(y)/(2)` |
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