A point P moves in a counter-clockwise direct | vedclass

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(A) Given the path length $S = t^3 + 5$.The tangential velocity $v$ is given by $v = \frac{dS}{dt} = 3t^2$.At $t = 2 \, s$,the velocity $v = 3(2)^2 =

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$14$

Vedclass · Answer

(A) Given the path length $S = t^3 + 5$.The tangential velocity $v$ is given by $v = \frac{dS}{dt} = 3t^2$.At $t = 2 \, s$,the velocity $v = 3(2)^2 = 12 \, m/s$.The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{R} = \frac{12^2}{20} = \frac{144}{20} = 7.2 \, m/s^2$.The tangential acceleration $a_t$ is given by $a_t = \frac{dv}{dt} = \frac{d}{dt}(3t^2) = 6t$.At $t = 2 \, s$,$a_t = 6(2) = 12 \, m/s^2$.The total acceleration $a$ is given by $a = \sqrt{a_c^2 + a_t^2} = \sqrt{(7.2)^2 + (12)^2} = \sqrt{51.84 + 144} = \sqrt{195.84} \approx 14 \, m/s^2$.

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