An object moves on a circular path; such that, its distance covered is given by the function $S = 2 t^{2}$ . The ratio of magnitudes of its accelerations at times $t_{1} = 1 s e c$ and $t_{2} = 2$ sec is 1 : 2. The radius of the circle is (Write until two digits after the decimal point)

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Solution

$v = \frac{d s}{d t} = 4 t$ & $a_{t} = \frac{d v}{d t} = 4$
$a = \sqrt{a_{t}^{2} + a_{c}^{2}} = \sqrt{16 + (\frac{v^{2}}{R})^{2}} = \sqrt{16 + \frac{16^{2} t^{4}}{R^{2}}}$
$\frac{a_{1}}{a_{2}} = \frac{1}{2} = \frac{\sqrt{16 + \frac{16^{2}}{R^{2}}}}{\sqrt{16 + 16^{2} \times \frac{16}{R^{2}}}} \Rightarrow 16 + \frac{16^{2} \times 16}{R^{2}} = (64 + \frac{16 t^{2}}{R^{2}}) \times 4$
$\frac{16^{2}}{R^{2}} (16 - 4) = 448 \Rightarrow R^{2} = \frac{16^{2}}{48} \times 12 = \frac{16^{2}}{2^{2}} \Rightarrow R = 8$

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An object moves on a circular path; such that, its distance covered is given by the function S=2t2. The ratio of magnitudes of its accelerations at times t1=1 sec and t2=2 sec is 1 : 2. The radius of the circle is (Write until two digits after the decimal point)

v=dsdt=4t & at=dvdt=4 a=√a2t+a2c=√16+(v2R)2=√16+162t4R2 a1a2=12=√16+162R2√16+162×16R2⇒16+162×16R2=(64+16t2R2)×4 162R2(16−4)=448⇒R2=16248×12=16222⇒R=8

An object moves on a circular path; such that, its distance covered is given by the function $S = 2 t^{2}$ . The ratio of magnitudes of its accelerations at times $t_{1} = 1 s e c$ and $t_{2} = 2$ sec is 1 : 2. The radius of the circle is (Write until two digits after the decimal point)