Question

A particle moves towards west with a velocity of \(10\,ms^{-1}\). After \(10\,s\) its direction changes towards south and it moves with the same velocity. The average acceleration of the particle is:

• A. \(2\sqrt{2}\,ms^{-2}\,NE\)
• B. \(\frac{1}{\sqrt{2}}\,ms^{-2}\,SE\)
• C. \(2\,ms^{-2}\,NE\)
• D. \(\sqrt{2}\,ms^{-2}\,SE\)

Hint:

Average acceleration depends on change in velocity, not path. Always use vector subtraction carefully.

Answer 1

The Correct Option is D

Step 1: Define initial and final velocities.
Initial velocity (west):
\[ \vec{v}_1 = -10\hat{i} \]
Final velocity (south):
\[ \vec{v}_2 = -10\hat{j} \]

Step 2: Change in velocity.
\[ \Delta \vec{v} = \vec{v}_2 - \vec{v}_1 \]
\[ \Delta \vec{v} = (-10\hat{j}) - (-10\hat{i}) = 10\hat{i} - 10\hat{j} \]

Step 3: Magnitude of change in velocity.
\[ |\Delta v| = \sqrt{(10)^2 + (-10)^2} \]
\[ |\Delta v| = \sqrt{200} = 10\sqrt{2} \]

Step 4: Time interval.
\[ \Delta t = 10\,s \]

Step 5: Average acceleration.
\[ a = \frac{\Delta v}{\Delta t} = \frac{10\sqrt{2}}{10} \]
\[ a = \sqrt{2}\,ms^{-2} \]

Step 6: Direction of acceleration.
Vector \(10\hat{i} - 10\hat{j}\) points towards south-east direction.

Step 7: Final conclusion.
\[ \boxed{\sqrt{2}\,ms^{-2}\,SE} \] Hence, correct answer is option (D).