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CSS Applied Mathematics QUESTION #1297
Question 1
For a particle moving along the ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) with uniform speed \(V\) at the point \((0,b)\), the normal component of acceleration is:
  • \(\dfrac{V^2 a}{b^2}\)✔️
  • \(\dfrac{V^2 b}{a^2}\)
  • \(\dfrac{V^2}{a}\)
  • \(\dfrac{V^2 a^2}{b^3}\)
Correct Answer Explanation
The radius of curvature of the ellipse at \((0,b)\) is \(\rho=\dfrac{b^2}{a}\) (standard result). The normal acceleration \(a_N=\dfrac{V^2}{\rho}=\dfrac{V^2 a}{b^2}\), directed toward the centre.