Question

The velocity and acceleration of a fluid particle are given as $\vec{V} = (-\hat{i} + 2\hat{j})$ m/s and $\vec{a} = (-2\hat{i} - 4\hat{j})$ m/s$^2$, respectively. The magnitude of the component of acceleration (in m/s$^2$, rounded off to two decimal places) of the fluid particle along the streamline is $________________$.

Hint:

For streamline acceleration, always project the acceleration vector onto the velocity direction using the dot product.

Answer 1

Correct Answer: 2.6

Step 1: Write velocity and acceleration vectors.
\[ \vec{V} = (-1, \ 2), \vec{a} = (-2, \ -4) \]
Step 2: Find magnitude of velocity.
\[ |\vec{V}| = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \approx 2.236 \]
Step 3: Component of acceleration along velocity direction.
\[ a_s = \frac{\vec{a} . \vec{V}}{|\vec{V}|} \] Dot product: \[ \vec{a} . \vec{V} = (-2)(-1) + (-4)(2) = 2 - 8 = -6 \] \[ a_s = \frac{-6}{2.236} \approx -2.683 \]
Step 4: Magnitude of component.
\[ |a_s| \approx 2.68 \ \text{m/s}^2 \] Final Answer: \[ \boxed{2.68 \ \text{m/s}^2} \]