Question

The coordinates of a moving particle are given by x = 4 -t²/2
and y =3+6t-t3/6

Find the velocity and Acceleration of the particle at t=2
sec.​

Answer 1

position of moving particle can be written as :-

$\rm \large \vec{r} = \bigg(4 - \dfrac{ {t}^{2} }{2} \bigg)\hat{i} + \bigg(3 + 6t - \dfrac{ {t}^{3} }{6} \bigg) \hat{j}$

Now, velocity = dr/dt

$\rm \rightarrow \large \dfrac{d\vec{r}}{dx} = \bigg(0 - \dfrac{2 {t} \ }{2} \bigg)\hat{i} + \bigg(0+ 6 - \dfrac{ 3{t}^{2} }{6} \bigg) \hat{j} \\ \\ \\ \rm\rightarrow \large \dfrac{d\vec{r}}{dx} = \bigg( - \dfrac{ {t} \ }{1} \bigg)\hat{i} + \bigg(6 - \dfrac{ {t}^{2} }{2} \bigg) \hat{j} \\ \\ \\ \rm \: put \: t = 2 \\ \\ \\ \rm\rightarrow \large v = \bigg( - {2 }{} \bigg)\hat{i} + \bigg(6 - \dfrac{ {2}^{2} }{2} \bigg) \hat{j}\\ \\ \\ \rm\rightarrow \large v = \bigg( - {2 }{} \bigg)\hat{i} + \bigg(6 - 2 \bigg) \hat{j}\\ \\ \\ \rm\rightarrow \large v = \bigg( - {2 }{} \bigg)\hat{i} + \bigg(4\bigg) \hat{j}\\ \\ \\ \rm\rightarrow \large v = \sqrt{ {( - 2)}^{2} + {(4)}^{2} } \\ \\ \\ \rm\rightarrow \large v = \sqrt{ 4+ 16 } \\ \\ \\ \rm\rightarrow \large v = \sqrt{ 20 }$

Now, acceleration = dv/dt

$\rm\rightarrow \large \dfrac{d\vec{v}}{dx} = \bigg( - 1 \bigg)\hat{i} + \bigg(0- \dfrac{ 2{t}^{} }{2} \bigg) \hat{j} \\ \\ \\ \rm\rightarrow \large \dfrac{d\vec{v}}{dx} = \bigg( - 1 \bigg)\hat{i} + \bigg(- \dfrac{ {t}^{} }{1} \bigg) \hat{j} \\ \\ \\ \rm \: put \: \: t = 2 \\ \\ \\ \rm\rightarrow \large \dfrac{d\vec{v}}{dx} = \bigg( - 1 \bigg)\hat{i} + \bigg(- { {2}^{} }\bigg) \hat{j} \\ \\ \\ \rm \: a = \sqrt{ {( - 1)}^{2} + {( - 2)}^{2} } \\ \\ \rm \: a = \sqrt{ 1 + 4} \\ \\ \rm \: a = \sqrt{ 5}$