Question

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 1

Answer:

The direction will be "39.8°". The further explanation is given below.

Explanation:

The equation will be:

⇒  $y=\frac{x^2}{24}$

On differentiating the above, we get

⇒  $\frac{dy}{dx}=\frac{2x}{24}$

         $=\frac{x}{12}$

On differentiating again, we get

⇒  $\frac{d^2y}{dx^2}=\frac{1}{12}$

Demonstrate the radius of the path curvature .

⇒  $\rho=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\left | \frac{d^2y}{dx^2} \right |}$

       $=\frac{[1+(\frac{x}{12} )^2]^{\frac{3}{2}}}{\left |\frac{1}{12} \right |}$

       $=\frac{[1+(\frac{10}{12})^2]^{\frac{3}{2}}}{\frac{1}{12} }$

       $=26.4 \ ft$

On calculating the acceleration's normal component, we get

⇒  $a_{n}=\frac{v^2}{\rho}$

         $=\frac{20}{26.4}$

         $=15.15 \ ft/s^2$

Magnitude,

⇒  $a=\sqrt{a_{n}^2+a_{t}^2}$

       $=\sqrt{(15.15)^2+(6)^2}$

       $=16.29 \ ft/s^2$

The direction of crate velocity will be:

⇒  $\phi=tan^{-1}(\frac{dy}{dx} )$

On putting the values, we get

       $=tan^{-1}(\frac{x}{12})$

       $=tan^{-1}(\frac{10}{12} )$

       $=39.8^{\circ}$