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In a two dimensional motion of a body, prove that tangentiol acceleration is nothing but component of acceleration along velocity. |
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Answer» `v=v_(x)hat(i)+v_(y)hat(j)` Accleration `a=(dv_(x))/(dt)hat(i)+(dv_(y))/(dt)hat(j)` Component of a along `v` will be, `(a.v)/(|v|)=(v_(x)(dv_(x))/(dt)+v_(y).(dv_(y))/(dt))/(sqrt(v_(x)^(2)+v_(y)^(y))` Further, tangential acceleration of particle is rate of change of speed. or `a_(t)=(dv)/(dt)=(d)/(dt)(sqrt(v_(x)^(2)+v_(y)^(2)))` or `a_(t)=(1)/(2sqrt(v_(x)^(2)+v_(y^(2))))[2v_(x).(dv_(x))/(dt)+2v_(y)(dv_(y))/(dt)]` or `a_(t)=(v_(x).(dv_(x))/(dt)+v_(y).(dv_(y))/(dt))/(sqrt(v_(x)^(2)+v_(y)^(2)))` Form Eqs. (i) and (ii), we can see that or Tangential acceleration=component of acceleration along velocity. |
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