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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane5 Marks
Question
A particle starts from the origin at t = 0s with a velocity of $10.0\hat{\text{j}}\text{m/s}$ and moves in the x - y plane with a constant acceleration of $(8.0\hat{\text{i}}+2.0\hat{\text{j}})\text{ms}^{-2}.$ (a) At what time is the x- coordinate of the particle 16m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at the time?
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Answer

Velocity of the particle, $\vec{\nu}=10.0\hat{\text{j}}\text{m/s}$ Acceleration of the particle, $\vec{\text{a}}=(8.0\hat{\text{i}}+2.0\hat{\text{j}})$ Also, But, $\vec{\text{a}}=\frac{\text{d}\vec{\nu}}{\text{dt}}=8.0\hat{\text{i}}+2.0\vec{\text{j}}$ $\text{d}\vec{\nu}=(8.0\hat{\text{i}}+2.0\hat{\text{j}})\text{dt}$ Integrating both sides: $\vec{\nu}(\text{t})=8.0\text{t}\hat{\text{i}}+2.0\text{t}\hat{\text{j}}+\vec{u}$ Where, $\vec{u}$ = Velocity vector of the particle at t = 0 $\vec{\nu}$ = Velocity vector of the particle at time t Integrating the equations with the conditions: at t = 0, r = 0, and at t = t, r = r $\vec{\text{r}}=\vec{u}\text{t}+\frac{1}{2}8.0\text{t}^2\hat{\text{i}}+\frac{1}{2}\times2.0\text{t}^2\hat{\text{j}}$ $=\vec{u}\text{t}+4.0\text{t}^2\hat{\text{i}}+\text{t}^2\hat{\text{j}}$ $=(10.0\hat{\text{j}})\text{t}+4.0\text{t}^2\hat{\text{i}}+\text{t}^2\hat{\text{j}}$ $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}=4.0\text{t}^2\hat{\text{i}}+(10\text{t}+\text{t}^2)\hat{\text{j}}$ We observe that the motion of the particle is in x - y plane, So, on equating the coefficients of $\hat{\text{i}}$ and $\hat{\text{j}}$ we get, $\text{x}=4\text{t}^2$ $\text{t}=\Big(\frac{\text{x}}{4}\Big)^{1/2}$ and $\text{y} = 10\text{t} + \text{t}^2$
  1. When y = 16m:
$\text{t}=\Big(\frac{16}{4}\Big)^{1/2}=2\text{s}$
$\therefore\text{y}=10\times2+(2)^2=24\text{m}$
  1. Velocity of the particle is given by:
$\vec{\nu}(\text{t})=8.0\text{t}\hat{\text{i}}+2.0\text{t}\hat{\text{j}}+\vec{u}$
$\text{At t}=2\text{s}$
$\vec{\nu}(2)=8.0\times2\hat{\text{i}}+2.0\times2\hat{\text{j}}+10\hat{\text{j}}$
$=16\hat{\text{i}}=14\hat{\text{j}}$
$\therefore$ Speed of the particle,
$\text{V}=\sqrt{(16)^2+(14)^2}$
$=\sqrt{256+196}=\sqrt{452}$
$=21.26\text{m/s}$

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