Respuesta :
Answer:
The velocity of the hay bale is - 0.5 ft/s and the acceleration is [tex]6.25\times 10^{- 3} ft/s^{2}[/tex]
Solution:
As per the question:
Constant velocity of the horse in the horizontal, [tex]v_{x} = 1 ft/s[/tex]
Distance of the horse on the horizontal axis, x = 10 ft
Vertical distance, y = 20 ft
Now,
Apply Pythagoras theorem to find the length:
[tex]20^{2} + 10^{2} = l^{2}[/tex]
[tex]l^{2}= 500[/tex]
Now,
[tex]x^{2} + y^{2} = 500[/tex] (1)
Differentiating equation (1) w.r.t 't':
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0[/tex]
[tex]x\frac{dx}{dt} = - y\frac{dy}{dt}[/tex]
where
[tex]\frac{dx}{dt}[/tex] = Rate of change of displacement along the horizontal
[tex]\frac{dy}{dt}[/tex] = Rate of change of displacement along the vertical
[tex]v_{x}[/tex] = velocity along the x-axis.
[tex]v_{y}[/tex] = velocity along the y-axis
[tex]xv_{x} = -yv_{y}[/tex]
[tex]v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s[/tex]
[tex]|v_{y}| = 0.5\ ft/s[/tex]
Acceleration of the hay bale is given by the kinematic equation:
[tex]v_{y}^{2} = u_{y} + 2ay[/tex]
[tex](-0.5)^{2} =0 + 2ay[/tex]
[tex]0.25 = 2ay[/tex]
[tex]\frac{0.25}{2y} = a[/tex]
[tex]a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}[/tex]
