When x = 10 ft, the crate has a speed of 20 ft/s which is increasing at 6 ft/s^2. Determine the direction of the crate's velocity and the magnitude of the crate's acceleration at this instant.

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]

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Engineering Design:Question 10
Which option describes the engineering design process?
Select one:
1. A series of steps engineers use to identify constraints.
2. A series of steps engineers use to decide which solution is best.
3. A series of steps engineers use to test what they build.
4. A series of steps engineers use to solve problems.