Question:
A particle moving in the x–y plane starts from the origin at \(t = 0\) with an initial velocity \((-\hat{i} + \hat{j}) \, \text{m/s}\) and undergoes an acceleration \((6\hat{i} + 4\hat{j}) \, \text{m/s}^2\). Its displacement after 2 s is:
A particle moving in the x–y plane starts from the origin at \(t = 0\) with an initial velocity \((-\hat{i} + \hat{j}) \, \text{m/s}\) and undergoes an acceleration \((6\hat{i} + 4\hat{j}) \, \text{m/s}^2\). Its displacement after 2 s is:
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For motion in 2D, always treat displacement, velocity, and acceleration as vectors; calculate each component separately and then use Pythagoras theorem to get magnitude.
Updated On: Jul 18, 2026
- 17.32 m
- 14.14 m
- 12.42 m
- 10 m
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The Correct Option is B
Solution and Explanation
Step 1: Use vector equation of motion.
Displacement \(\vec{s}\) in 2D under constant acceleration is given by:
\[ \vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^2 \]
where \(\vec{u} = -\hat{i} + \hat{j}\) m/s, \(\vec{a} = 6\hat{i} + 4\hat{j}\) m/s\(^2\), and \(t = 2\) s.
Step 2: Compute displacement components.
- x-component:
\[ s_x = u_x t + \frac{1}{2} a_x t^2 = (-1)(2) + \frac{1}{2}(6)(2^2) = -2 + 12 = 10 \, \text{m} \]
- y-component:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 = (1)(2) + \frac{1}{2}(4)(2^2) = 2 + 8 = 10 \, \text{m} \]
Step 3: Calculate magnitude of displacement.
\[ |\vec{s}| = \sqrt{s_x^2 + s_y^2} = \sqrt{10^2 + 10^2} = \sqrt{200} \]
Step 4: Simplify the magnitude.
\[ |\vec{s}| = \sqrt{200} = 14.14 \, \text{m} \]
Step 5: Verify units and correctness.
All components are in meters, time in seconds, and acceleration in m/s\(^2\). Calculation of vector magnitude correctly gives displacement.
Step 6: Final conclusion.
Hence, the particle's displacement after 2 s is:
\[ \boxed{14.14 \, \text{m}} \]
Displacement \(\vec{s}\) in 2D under constant acceleration is given by:
\[ \vec{s} = \vec{u} t + \frac{1}{2} \vec{a} t^2 \]
where \(\vec{u} = -\hat{i} + \hat{j}\) m/s, \(\vec{a} = 6\hat{i} + 4\hat{j}\) m/s\(^2\), and \(t = 2\) s.
Step 2: Compute displacement components.
- x-component:
\[ s_x = u_x t + \frac{1}{2} a_x t^2 = (-1)(2) + \frac{1}{2}(6)(2^2) = -2 + 12 = 10 \, \text{m} \]
- y-component:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 = (1)(2) + \frac{1}{2}(4)(2^2) = 2 + 8 = 10 \, \text{m} \]
Step 3: Calculate magnitude of displacement.
\[ |\vec{s}| = \sqrt{s_x^2 + s_y^2} = \sqrt{10^2 + 10^2} = \sqrt{200} \]
Step 4: Simplify the magnitude.
\[ |\vec{s}| = \sqrt{200} = 14.14 \, \text{m} \]
Step 5: Verify units and correctness.
All components are in meters, time in seconds, and acceleration in m/s\(^2\). Calculation of vector magnitude correctly gives displacement.
Step 6: Final conclusion.
Hence, the particle's displacement after 2 s is:
\[ \boxed{14.14 \, \text{m}} \]
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