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The ends of a rod AB which is 5 m long m...

The ends of a rod AB which is 5 m long moves along two grooves `OX, OY` which are at right angles. If A moves 2 at a constant speed of `1/2m/s`, what is the speed of B, when it is 4 m from O?

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To solve the problem, we will follow these steps: ### Step 1: Understand the Geometry We have a rod AB of length 5 m, where point A moves along the x-axis (OX) and point B moves along the y-axis (OY). The coordinates of point A can be represented as (x, 0) and the coordinates of point B as (0, y). Since the rod is of fixed length, we can use the Pythagorean theorem to relate x and y. ### Step 2: Set Up the Equation Using the Pythagorean theorem, we have: \[ x^2 + y^2 = 5^2 \] This simplifies to: \[ x^2 + y^2 = 25 \] ### Step 3: Differentiate with Respect to Time To find the relationship between the rates of change of x and y, we differentiate both sides of the equation with respect to time \( t \): \[ \frac{d}{dt}(x^2) + \frac{d}{dt}(y^2) = \frac{d}{dt}(25) \] Using the chain rule, we get: \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] Dividing through by 2 gives: \[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \] ### Step 4: Substitute Known Values We know: - The speed of point A, \( \frac{dx}{dt} = v_A = \frac{1}{2} \) m/s (moving away from O). - We need to find \( \frac{dy}{dt} = v_B \) when point B is 4 m from O, which means \( y = 4 \). From the Pythagorean theorem, when \( y = 4 \): \[ x^2 + 4^2 = 25 \implies x^2 + 16 = 25 \implies x^2 = 9 \implies x = 3 \text{ m} \] ### Step 5: Plug Values into the Differentiated Equation Substituting \( x = 3 \), \( y = 4 \), and \( \frac{dx}{dt} = \frac{1}{2} \) into the differentiated equation: \[ 3 \left(\frac{1}{2}\right) + 4 \frac{dy}{dt} = 0 \] This simplifies to: \[ \frac{3}{2} + 4v_B = 0 \] ### Step 6: Solve for \( v_B \) Rearranging gives: \[ 4v_B = -\frac{3}{2} \implies v_B = -\frac{3}{8} \text{ m/s} \] ### Conclusion The speed of point B when it is 4 m from O is: \[ v_B = -\frac{3}{8} \text{ m/s} \] The negative sign indicates that point B is moving downward along the y-axis.
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