The formula for the speed of the wagon can be found by using Pythagoras theorem, and chain rule of differentiation.
[tex](a) \ \mathrm{The \ speed \ of \ the \ wagon} , \ \dfrac{dx}{dt} = q -\dfrac{x}{\sqrt{x^2+ 5.76}}[/tex]
(b) When x = 0.6 and q = 0.5 m/s, the speed of the wagon is approximately 0.243 m/s.
Reasons:
(a) The distance from the cart to the pulley, r is given by Pythagoras's
theorem as follows;
r = √((3 - 0.6)² + x²) = √(2.4² + x²)
The speed of the wagon = [tex]\mathbf{\dfrac{dx}{dt}}[/tex]
The speed of the rope, q = [tex]\mathbf{\dfrac{dr}{dt}}[/tex]
By chain rule, we have;
[tex]\dfrac{dr}{dt} = \mathbf{\dfrac{dr}{dx} + \dfrac{dx}{dt}}[/tex]
[tex]\dfrac{dr}{dx} = \dfrac{x}{\sqrt{x^2+ 5.76}}[/tex]
Therefore;
[tex]\dfrac{dr}{dt} = q =\dfrac{x}{\sqrt{x^2+ 5.76}}+ \dfrac{dx}{dt}[/tex]
[tex]\mathrm{The \ speed \ of \ the \ wagon} , \ \dfrac{dx}{dt} = \underline{ q -\dfrac{x}{\sqrt{x^2+ 5.76}}}[/tex]
(b) The speed of the wagon when x = 0.6 if q = 0.5 m/s is given as follows;
[tex]\dfrac{dx}{dt} = 0.5 -\dfrac{0.6}{\sqrt{0.6^2+ 5.76}} =\sqrt{\dfrac{1}{17} } \approx 0.243[/tex]
The speed of the wagon when x = 0.6 and q = 0.5 m/s, [tex]\dfrac{dx}{dt}[/tex] ≈ 0.243 m/s
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