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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane5 Marks
Question
The position of a particle is given by $\text{r}=3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{j}}+4.0\hat{\text{k }}\text{m}$ Where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0s?
✓

Answer

  1. $\vec{\nu}(\text{t})=(3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}});\text{a}=-4.0\hat{\text{j}}$
The position of the particle is given by:
$\vec{\text{r}}=3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{J}}+4.0\hat{\text{k}}$
Velocity $\vec{\nu},$ of the particle given as:
$\vec{\nu}=\frac{\text{d}\vec{\text{r}}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(3.0\text{t}\hat{\text{i}}-2.0\text{t}^2\hat{\text{J}}+4.0\hat{\text{k}})$
$\therefore\vec{\nu}=3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}}$
Acceleration, $\vec{\text{a}},$ of the particle is given as:
$\vec{\text{a}}=\frac{\text{d}\vec{\nu}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}})$
$\therefore\vec{\text{a}}=-4.0\hat{\text{J}}$
  1. 8.54m/s, 69.45° below the x - axis
We have velocity vector, $\vec{\nu}=3.0\hat{\text{i}}-4.0\text{t}\hat{\text{J}}$
$\text{At t}=2.0\text{s}:$
$\vec{\nu}=3.0\hat{\text{i}}-8.0\hat{\text{J}}$
The magnitude of veocity is given by:
$|\vec{\nu}|\sqrt{3^2+(-8)^2}=\sqrt{73}=8.54\text{m/s}$
Direction, $\theta=\tan^{-1}\Big(\frac{\nu_\text{y}}{\nu_\text{x}}\Big)$
$=\tan^{-1}\Big(\frac{-8}{3}\Big)=-\tan^{-1}(2.667)$
$=-69.45^\circ$
The negative sing indicates that the direction of velocity is below the x - axis.

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