The positon vector → r of a particle of mass m is given by the...

\overset{\to}{r} = \alpha{t}^{3}\overset{^}{i} + \beta{t}^{2}\overset{^}{j}$<br >$\overset{\to}{v} = \dfrac{d\overset{\to}{r}}{dt} = 3\alpha{t}^{2}\overset{^}{i} + 2\betat\overset{^}{j}$<br >$\overset{\to}{a} = \dfrac{{d}^{2}\overset{\to}{r}}{d{t}^{2}} = 6\alphat\overset{^}{i} + 2\beta\overset{^}{j}$<br >At t = 1<br >(i)$\overset{\to}{v} = 3 \times \dfrac{10}{3} \times 1\overset{^}{i} + 2 \times J \times 1\overset{^}{j}$<br >$= 10\overset{^}{i} + 10\overset{^}{j}$<br >(ii)$\overset{\to}{L} = \overset{\to}{r} \times \overset{\to}{p}<br > =\left(\dfrac{10}{3} \times 1\overset{^}{i} + 5 \times 1\overset{^}{j}\right) \times 0.1\left(10\overset{^}{i} + 10\overset{^}{j}\right)
= - \dfrac{5}{3}\overset{^}{k}
(iii) \overset{\to}{F} = m \times \left(6 \times \dfrac{10}{3} \times 1\overset{^}{i} + 2 \times 5\overset{^}{j}\right) = 2\overset{^}{i} + \overset{^}{j}
(iv) $\overset{\to}{τ} = r \times \overset{\to}{F}$
=\left(\dfrac{10}{3}\overset{^}{i} + 5\overset{^}{j}\right) \times \left(2\overset{^}{i} + \overset{^}{j}\right)$<br >$= + \dfrac{10}{3}\overset{^}{k} + 10\left(- \overset{^}{k}\right)
= - \dfrac{20}{3}\overset{^}{k}