Q180 / 180 · KCET 2025 · Physics
A particle is in uniform circular motion. The equation of its trajectory is given by $(x-2)^2+y^2=25$, where x and y are in meter. The speed of the particle is $2 \mathrm{~ms}^{-1}$, when the particle attains the lowest ' $y$ ' co-ordinate, the acceleration of the particle is (in $\mathrm{ms}^{-2}$ )
- A$0.4 \hat{\mathrm{j}}$
- B$0.8 \hat{\mathrm{i}}$
- Correct answerC$0.8 \hat{\mathrm{j}}$
- D$0.4 \hat{\mathrm{i}}$
Solution
$(x-2)^2+y^2=25$
Equation of circle center $\Rightarrow(2,0)$
$\begin{aligned}
& r^2=25 \\
& r=5 \\
& a_c=\frac{v^2}{R}=\frac{2^2}{5} \\
& =0.8 \mathrm{~m} / \mathrm{s}^2 \\
& \overrightarrow{\mathrm{a}_{\mathrm{c}}}=0.8 \hat{\mathrm{i}} \mathrm{~m} / \mathrm{s}^2
\end{aligned}$
Hence, the correct answer is option: 'C'.