Question

A particle starts from origin at \( t = 0 \) with a velocity \( 5\hat{i} \, \text{m/s} \) and moves in the \( x\)-\( y \) plane under the action of a force that produces a constant acceleration of \( (3\hat{i} + 2\hat{j}) \, \text{m/s}^2 \). If the \( x \)-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is \( \sqrt{\alpha} \, \text{m/s} \). The value of \( \alpha \) is ______.

Answer 1

Correct Answer: 673

Given initial velocity \( u_x = 5 \, \text{m/s} \), acceleration \( a_x = 3 \, \text{m/s}^2 \), and \( x = 84 \, \text{m} \).

Using the equation:

\[ v_x^2 = u_x^2 + 2a_xx \]

\[ v_x^2 = 5^2 + 2 \cdot 3 \cdot 84 = 25 + 504 = 529 \]

\[ v_x = 23 \, \text{m/s} \]

Similarly, for the y-direction:

\[ v_y = u_y + a_y t = 0 + 2 \cdot t \]

Using \( v_x = u_x + a_x t \):

\[ t = \frac{v_x - u_x}{a_x} = \frac{23 - 5}{3} = 6 \, \text{s} \]

Then,

\[ v_y = 2 \times 6 = 12 \, \text{m/s} \]

The speed of the particle is:

\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \]

Thus, \(\alpha = 673\).