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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane3 Marks
Question
The position of a particle is given by $\vec{\text{r}}=9\text{t}\hat{\text{i}}+6\text{t}^2\hat{\text{j}}+8\hat{\text{k}},$ where $t$ is in seconds and the coefficients have the proper units for $\vec{\text{r}}$ to be in metres.
  1. Find velocity $\vec{\text{v}}(\text{t})$ of particle and $\vec{\text{a}}(\text{t})$ acceleration of the particle.
  2. Find the magnitude and direction of $\vec{\text{v}}(\text{t})$ at $t = 2sec.$
✓

Answer

$\vec{\text{r}}=9\text{t}\hat{\text{i}}+6\text{t}^2\hat{\text{j}}+8\hat{\text{k}}$ $\vec{\text{v}}(\text{t})=\frac{\vec{\text{dr}}}{\text{dt}}=9\hat{\text{i}}+12\text{t}\hat{\text{j}}$ $\vec{\text{a}}(\text{t})=\frac{\vec{\text{dv}}}{\text{dt}}=12\hat{\text{j}} = 12m/ s$ along $y-$axis At $t = 2 \vec{\text{v}}=9\hat{\text{i}}+12(2)\hat{\text{j}}$
or, $\vec{\text{v}}=9\hat{\text{i}}+24\hat{\text{j}}$
magnitude, $|\vec{\text{v}}|=\sqrt{(9)^2+(24)^2}=25\text{ m/s}$ direction $\theta=\tan^{-1}\Big(\frac{25}{9}\Big)$ with $x-$axis

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