Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

56 m

64 m

80 m

128 m

Step-by-Step Solution

Identify Given Values: Initial velocity, $\mathbf{u} = 0$ (starts from rest). Acceleration vector, $\mathbf{a} = a_x \hat{i} + a_y \hat{j} = 6\hat{i} + 8\hat{j} \text{ m/s}^2$ .

Time, $t = 4 \text{ s}$ .

Calculate Net Acceleration: The magnitude of the resultant acceleration is $a = \sqrt{a_x^2 + a_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m/s}^2$ .
Calculate Distance: Since the body starts from rest with constant acceleration, the distance covered ( $s$ ) is given by the kinematic equation $s = ut + \frac{1}{2}at^2$ . $s = 0 + \frac{1}{2}(10)(4)^2$ $s = 5 \times 16 = 80 \text{ m}$ .
Alternative Method (Component-wise): $x$ -distance: $x = \frac{1}{2} a_x t^2 = \frac{1}{2}(6)(16) = 48 \text{ m}$ . $y$ -distance: $y = \frac{1}{2} a_y t^2 = \frac{1}{2}(8)(16) = 64 \text{ m}$ .

Total distance from origin: $r = \sqrt{x^2 + y^2} = \sqrt{48^2 + 64^2} = 80 \text{ m}$ .

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