(A) Given,the path length $s = t^3 + 5$.Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(t^3 + 5) = 3t^2 \ m/s$.Tangential acceleration $a_t = \frac{dv}{dt}

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(A) Given,the path length $s = t^3 + 5$.Velocity $v = \frac{ds}{dt} = \frac{d}{dt}(t^3 + 5) = 3t^2 \ m/s$.Tangential acceleration $a_t = \frac{dv}{dt} = \frac{d}{dt}(3t^2) = 6t \ m/s^2$.Radial (centripetal) acceleration $a_c = \frac{v^2}{R} = \frac{(3t^2)^2}{R} = \frac{9t^4}{R} \ m/s^2$.At $t = 2 \ s$:$a_t = 6 \times 2 = 12 \ m/s^2$.$a_c = \frac{9 \times (2)^4}{20} = \frac{9 \times 16}{20} = \frac{144}{20} = 7.2 \ m/s^2$.The resultant acceleration $a = \sqrt{a_t^2 + a_c^2}$.$a = \sqrt{(12)^2 + (7.2)^2} = \sqrt{144 + 51.84} = \sqrt{195.84} \approx 14 \ m/s^2$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ point $P$ moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length $s = t^3 + 5$,where $s$ is in meters and $t$ is in seconds. The radius of the path is $20 \ m$. The acceleration of $P$ when $t = 2 \ s$ is nearly .......... $m/s^2$.

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A
$14$
B
$13$
C
$12$
D
$7.2$

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3-2.Motion in Plane

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Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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