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3-2.Motion in Plane
medium
The position of a particle moving in the $xy-$plane at any time $t$ is given by $x = (3{t^2} - 6t)$ metres, $y = ({t^2} - 2t)$ metres. Select the correct statement about the moving particle from the following
A
The acceleration of the particle is zero at $t = 0$ second
B
The velocity of the particle is zero at $t = 0$ second
C
The velocity of the particle is zero at $t = 1$ second
D
The velocity and acceleration of the particle are never zero
Solution
(c) ${v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}(3{t^2} – 6t) = 6t – 6$.
At $t = 1,\;{v_x} = 0$
${v_y} = \frac{{dy}}{{dt}} = \frac{d}{{dt}}({t^2} – 2t) = 2t – 2$.
At $t = 1,\;{v_y} = 0$
Hence $v = \sqrt {v_x^2 + v_y^2} = 0$
Standard 11
Physics
