$\vec{a}=2 \hat{i}+2 t \hat{j}$
$|\vec{V}|=V=\sqrt{4 t^{2}+t^{4}}$
$a_{t}=\left(\frac{d V}{d t}\right)=\frac{1}{2 \sqrt{4 t^{2}+t^{4}}} \times\left(8 t+4 t^{3}\right)$
at $t=1 \quad a_{t}=\left(\frac{6}{\sqrt{5}}\right) \mathrm{m} / \mathrm{s}^{2}$
$a=\sqrt{2^{2}+(2 t)^{2}}=\sqrt{8} \mathrm{m} / \mathrm{s}^{2}$
$r=\sqrt{5}$
$a_{c}=\sqrt{a^{2}-a_{t}^{2}}=\sqrt{8-\frac{36}{5}}=\left(\frac{2}{\sqrt{5}}\right)$
$R=\left(\frac{V^{2}}{a_{c}}\right)=\frac{5}{\left(\frac{2}{\sqrt{5}}\right)}=\frac{5 \sqrt{5}}{2}$
$R=\left(\frac{5 \sqrt{5}}{2}\right) m$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(1)$ a ring
$(2)$ a disc
$(3)$ a solid cylinder
$(4)$ a solid sphere,
of same mass $m$ and radius $R$ are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is ...........
[Mark the body as per their respective numbering given in the question]
