Question

The position vector is given by r=5t^2icap +2t^3jcap+2kcap. Find its velocity and acceleration at t=2s

Answer 1

Explanation:

We have,

$\vec{r} =( 5 {t}^{2} )\hat{i} + (2 {t}^{3} ) \hat{j} + (2) \hat{k}$

Its velocity is given by

$\vec{v} = \frac{d \vec{r}}{dt} = \bigg \{ \frac{d}{dt} (5 {t}^{2}) \bigg \} \hat{i} + \bigg \{ \frac{d}{dt} (2 {t}^{3}) \bigg \} \hat{j} +\bigg \{ \frac{d}{dt} (2) \bigg \} \hat{k}$

$\implies \vec{v} = (10t) \hat{i} + (6 {t}^{2}) \hat{j} + (0)\hat{k}$

$\implies \vec{v} = (10t) \hat{i} + (6 {t}^{2}) \hat{j}$

So, its velocit at $t=2s$ is,

$\implies \vec{v} = (10 \times 2) \hat{i} + (6 \times {(2)}^{2}) \hat{j}$

$\implies \vec{v} = 20 \hat{i} + 24 \hat{j}$

Now,

$\vec{a} = \frac{d \vec{v}}{dt} = \bigg \{ \frac{d}{dt} (10t) \bigg \} \hat{i} + \bigg \{ \frac{d}{dt} (6 {t}^{2}) \bigg \} \hat{j}$

$\implies \vec{a} = (10) \hat{i} + (12 t) \hat{j}$

So, its acceleration at $t=2s$,

$\implies \vec{a} = (10) \hat{i} + (12 (2)) \hat{j}$

$\implies \vec{a} = 10 \hat{i} + 24 \hat{j}$