A particle starts from the origin at t = 0 s with the velocity of \(10 \; \hat{j} \; m \; s^{-1}\) and moves in the x –y plane with a constant acceleration of** \(\left ( 8.0 \; \hat{i} + 2.0 \; \hat{j}\right ) \; m \; s^{-2}\)**

**(a)** At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

**(b)** What is the speed of the particle at the time?

Asked by Abhisek | 1 year ago | 80

**(a) Velocity of the particle** =\( 10 \; \hat{j} \; m \; s^{-1}\)

Acceleration of the particle = \(\left (8.0 \; \hat{i} + 2.0 \; \hat{j}\right) \; m \; s^{-2}\)

But, \(\vec{a} = \dfrac{\vec{dv}}{dt} = 8.0 \; \hat{i} + 2.0 \; \hat{j}\vec{dv} = \left (8.0 \; \hat{i} + 2.0 \; \hat{j} \right )\ ;dt\)

**Integrating both the sides:**

\(\vec{v}\left ( t \right ) = 8.0 t \; \hat{i} + 2.0 t \; \hat{j} + \vec{u}\)

Where,\( \vec{u}\) = velocity vector of the particle at t = 0

\( \vec{v}\)= velocity vector of the particle at time t

But, \(\vec{v} = \dfrac{\vec{dr}}{dt}[/latex]\vec{dr} = \vec{v}\; dt\)

=\( \left (8.0 t \; \hat{i} + 2.0 t \; \hat{j} + \vec{u} \right ) \; dt\)

**Integrating the equations with the conditions:**

At t = 0; r = 0 and at t = t; r = r.

\( \vec{r} = \vec{u}t + \frac{1}{2}8.0t^{2} \; \hat{i} + \frac{1}{2} \times 2.0t^{2} \; \hat{j}\vec{r} \)

\( = \vec{u}t + 4.0t^{2} \; \hat{i} + t^{2} \; \hat{j}\vec{r} \)

\( = \left (10.0 \; \hat{j} \right)t + 4.0t^{2} \; \hat{i} + t^{2} \; \hat{j}\times \;\hat{i} + y \;\hat{j} \)

\( = 4.0t^{2} \; \hat{i} + \left ( 10.0 \; t + t^{2} \right )\; \hat{j}\)

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of

\(\hat{i} \; and \; \hat{j}\), we get:

\(x = 4t^{2}\)

t =\(\left (\dfrac{x}{4} \right)^{\dfrac{1}{2}}\)

y = \(10 t + t^{2}\)

**When x = 16 m:**

\(t = \left (\dfrac{16}{4} \right )^{\dfrac{1}{2}}\) = 2 s

Therefore, y = 10 × 2 + \(\left (2 \right)^{2}\) **= 24 m**

**(b) Velocity of the particle:**

\(\vec{v} ( t ) = 8.0 t \; \hat{i} + 2.0 t \; \hat{j} + \vec{u}\)

**At t = 2 s:**

\(\vec{v} ( t ) = 8.0 \times 2 \; \hat{i} + 2.0 \times 2 \; \hat{j} + 20 \; \hat{j}\vec{v} ( t ) = 16 \; \hat{i} + 14\; \hat{j}\)

**Therefore, Speed of the particle:**

\(\left |\vec{v} \right | = \sqrt{\left ( 16 \right )^{2} + \left ( 14 \right )^{2}}\left |\vec{v} \right | = \sqrt{256 + 196}\left |\vec{v} \right | = \sqrt{452}\left |\vec{v} \right | = 21.26 \; m \; s^{-1}\)

Answered by Pragya Singh | 1 year ago**(a) **Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

\(\theta (t)=tan^{-1}(\dfrac{v_{0y-gt}}{v_{ox}})\)

**(b)** Shows that the projection angle θ_{0} for a projectile launched from the origin is given by

\(\theta_{0}=tan^{-1}(\dfrac{4h_{m}}{R})\)

where the symbols have their usual meaning

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