A body starts from rest from the origin with an acceleration of $6 ext{ m/s}^2$ along the $x$-axis and $8

Question

A body starts from rest from the origin with an acceleration of $6 	ext{ m/s}^2$ along the $x$-axis and $8 	ext{ m/s}^2$ along the $y$-axis. Its distance from the origin after $4 	ext{ seconds}$ will be:

Accepted Answer

1.  **Identify Given Values:**
    *   Initial velocity, $\mathbf{u} = 0$ (starts from rest).
    *   Acceleration vector, $\mathbf{a} = a_x \hat{i} + a_y \hat{j} = 6\hat{i} + 8\hat{j} 	ext{ m/s}^2$.
    *   Time, $t = 4 	ext{ s}$.
2.  **Calculate Net Acceleration:** The magnitude of the resultant acceleration is $a = \sqrt{a_x^2 + a_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 	ext{ m/s}^2$ .
3.  **Calculate Distance:** Since the body starts from rest with constant acceleration, the distance covered ($s$) is given by the kinematic equation $s = ut + \frac{1}{2}at^2$ .
    $s = 0 + \frac{1}{2}(10)(4)^2$
    $s = 5 	imes 16 = 80 	ext{ m}$.
4.  **Alternative Method (Component-wise):**
    *   $x$-distance: $x = \frac{1}{2} a_x t^2 = \frac{1}{2}(6)(16) = 48 	ext{ m}$.
    *   $y$-distance: $y = \frac{1}{2} a_y t^2 = \frac{1}{2}(8)(16) = 64 	ext{ m}$.
    *   Total distance from origin: $r = \sqrt{x^2 + y^2} = \sqrt{48^2 + 64^2} = 80 	ext{ m}$.

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