Answer:
[tex]\frac{dy}{dt}=-\frac{2(7+2t)}{\sqrt{25^2-(7+2t)^2}}\\[/tex]
for y (0,9)s
Explanation:
the average velocity can be computed by using the derivative:
[tex]\frac{dy}{dt}[/tex]
For this case we have
[tex]y(t)=\sqrt{25^2-(7+2t)^2}=(25^2-(7+2t)^2)^{\frac{1}{2}}\\\\\frac{dy}{dt}=\frac{1}{2}(25^2-(7+2t)^2)^{-\frac{1}{2}}(-2(7+2t)(2))\\\\\frac{dy}{dt}=-\frac{2(7+2t)}{\sqrt{25^2-(7+2t)^2}}[/tex]
However, for a real result the values of t must be
[tex]7+2t\leq 25\\t\leq 9[/tex]
that is, an allowed interval for t is (0,9)
hope this helps!!
Answer:
[tex]V_{B} = -2*\frac{\frac{Xo + 2t}{L} }{\sqrt{1-(\frac{Xo + 2t}{L}) ^{2} } } \n [m/s][/tex]
Explanation:
Since:
[tex]Y_{B} = L*sin(\alpha )[/tex]
[tex]cos(\alpha ) = \frac{Xo + 2t}{L} ---> \alpha = arcCos(\frac{Xo + 2t}{L})[/tex] ---(1)
By compositive derivate:
d(YB)/dt =VB = L*(d[sin(α)]/dα)*(dα/dt) = L*cos(α)*dα/dt --- (2)
From the expresion (1):
if we call; u = (Xo + 2t)/L
Then: dα/dt = (dα/du)*(du/dt) = (d[arcCos(u)]/du)*(d[(Xo + 2t)/L]/dt)
Thus:
[tex]d\alpha /dt = -\frac{1}{\sqrt{1-u^{2} } } *(\frac{2}{L}) = -\frac{1}{\sqrt{1-cos\alpha ^{2} } }*(\frac{2}{L})[/tex]
[tex]d\alpha /dt = -\frac{1}{\sin\alpha } *(\frac{2}{L})[/tex]
Finally:
[tex]V_{B} = L*cos\alpha *(-\frac{1}{sin\alpha }) *\frac{2}{L} = -2*Ctg\alpha [m/s][/tex]
Or, in cartesians:
[tex]V_{B} = -2*\frac{cos\alpha }{\sqrt{1-cos\alpha ^{2} } } [m/s][/tex]
[tex]V_{B} = -2*\frac{\frac{Xo + 2t}{L} }{\sqrt{1-(\frac{Xo + 2t}{L}) ^{2} } } \n [m/s][/tex]
