The distance of a particle moving on a circle | vedclass

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(B) Given,$s = 2t^3$ and radius $R = 12 \, m$.The velocity $v$ is the rate of change of distance along the path: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^

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$1: 2$

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(B) Given,$s = 2t^3$ and radius $R = 12 \, m$.The velocity $v$ is the rate of change of distance along the path: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3) = 6t^2$.At $t = 2 \, s$,the velocity $v = 6(2)^2 = 24 \, m/s$.The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R} = \frac{24^2}{12} = \frac{576}{12} = 48 \, m/s^2$.The tangential acceleration $a_t$ is the rate of change of speed: $a_t = \frac{dv}{dt} = \frac{d}{dt}(6t^2) = 12t$.At $t = 2 \, s$,the tangential acceleration $a_t = 12(2) = 24 \, m/s^2$.The ratio of tangential to centripetal acceleration is $\frac{a_t}{a_c} = \frac{24}{48} = \frac{1}{2}$,which is $1: 2$.

The distance of a particle moving on a circle of radius $12 \, m$ measured from a fixed point on the circle and measured along the circle is given by $s = 2t^3$ (in meters). The ratio of its tangential to centripetal acceleration at $t = 2 \, s$ is .........

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