Respuesta :
Answer:
The direction will be "39.8°". The further explanation is given below.
Explanation:
The equation will be:
⇒ [tex]y=\frac{x^2}{24}[/tex]
On differentiating the above, we get
⇒ [tex]\frac{dy}{dx}=\frac{2x}{24}[/tex]
[tex]=\frac{x}{12}[/tex]
On differentiating again, we get
⇒ [tex]\frac{d^2y}{dx^2}=\frac{1}{12}[/tex]
Demonstrate the radius of the path curvature .
⇒ [tex]\rho=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\left | \frac{d^2y}{dx^2} \right |}[/tex]
[tex]=\frac{[1+(\frac{x}{12} )^2]^{\frac{3}{2}}}{\left |\frac{1}{12} \right |}[/tex]
[tex]=\frac{[1+(\frac{10}{12})^2]^{\frac{3}{2}}}{\frac{1}{12} }[/tex]
[tex]=26.4 \ ft[/tex]
On calculating the acceleration's normal component, we get
⇒ [tex]a_{n}=\frac{v^2}{\rho}[/tex]
[tex]=\frac{20}{26.4}[/tex]
[tex]=15.15 \ ft/s^2[/tex]
Magnitude,
⇒ [tex]a=\sqrt{a_{n}^2+a_{t}^2}[/tex]
[tex]=\sqrt{(15.15)^2+(6)^2}[/tex]
[tex]=16.29 \ ft/s^2[/tex]
The direction of crate velocity will be:
⇒ [tex]\phi=tan^{-1}(\frac{dy}{dx} )[/tex]
On putting the values, we get
[tex]=tan^{-1}(\frac{x}{12})[/tex]
[tex]=tan^{-1}(\frac{10}{12} )[/tex]
[tex]=39.8^{\circ}[/tex]
