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3-2.Motion in Plane
normal
A point $P$ moves in counter clock wise direction on a circular path as shown in figure. The movement of $'P'$ is such that it sweeps out a length $S = t^3 + 5$, where $'S'$ is in meter and $t$ is in seconds. The radius of the path is $20\, m$. The acceleration of $'P'$ when $t = 2\, sec$. is nearly ......... $m/s^2$
A
$14$
B
$13$
C
$12$
D
$7.2$
Solution
$S=t^{3}+5$
$\mathrm{V}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^{2}$
$\mathrm{V}_{2}=3(2)^{2}=12 \mathrm{m} / \mathrm{s}$
$a_{c}=\frac{v^{2}}{r}=\frac{12 \times 12}{20}=7.2 \mathrm{m} / \mathrm{s}^{2}$
$a_{t}=\frac{d^{2} s}{d t^{2}}=6 t \quad \Rightarrow \quad\left(a_{t}\right)_{t=2}=12 m / s^{2}$
$a=a_{c}^{2}+a t_{t}^{2}=14 \mathrm{m} / \mathrm{s}^{2}$
Standard 11
Physics
