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The co-ordinates of a moving particle at any time t are given by `x=ct^(2) and y=bt^(2)`The speed of the particle isA. `2t(c+b)`B. `2tsqrt(c^(2)-b^(2))`C. `tsqrt(c^(2)+b^(2))`D. `2t sqrt(c^(2)+b^(2))` |
Answer» Correct Answer - D `|vec(v)|=sqrt(v_(x)^(2)+v_(y)^(2))` here `v_(x)=(dx)/(dt)=2ct, v_(y)=(dy)/(dt)=2bt` therefore `|vec(v)|=sqrt(4t^(2)(c^(2)+b^(2)))=2tsqrt((c^(2)+b^(2)))` |
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