Answer to Question #128860 in Mechanics | Relativity for Ariyo Emmanuel

Question

When x = 10 ft, the crate has a speed of 20 ft/s which is increasing at 6 ft/s2. Determine the direction of the crate’s velocity and the magnitude of the crate’s acceleration at this instant.

Answer 1

To answer this question we should know the equation of the trajectory of the crate.

Let us consider the trajectory given in this problem as an example: https://www.chegg.com/homework-help/questions-and-answers/3-x-10-ft-crate-speed-20-ft-s-increasing-6-ft-s-2-determine-direction-crate-s-velocity-mag-q8097639

The velocity is directed along the tangent line to the trajectory. Let us calculate the coordinates of the point: $\,\, x_0=10\,\text{ft},\;\; y_0 = \dfrac{1}{24}x^2 = \dfrac{100}{24} = \dfrac{25}{6}\,\text{ft}.$

The tangent line is tilted at an angle with $\tan \alpha = y'(x_0) = \dfrac{1}{12}\cdot10 \approx 0.833, \; \alpha \approx 39.8^\circ.$ The direction of the velocity vector makes an angle $-39.8^\circ$ with the x-axis.

Let us determine the curvature of the trajectory in point in question:

$\kappa = \dfrac{|f''|}{(\sqrt{1+f'^2})^3} = \dfrac{\dfrac{1}{12}}{\left(\sqrt{1+\dfrac{25}{36}}\right)^3} = \dfrac{125}{12\sqrt{61^3}} \approx 0.0378,$

so the radius of curvature is $r=1/\kappa \approx 26.5\,\text{ft}.$

We know the tangential component of the acceleration (6 ft/s^2) and we should calculate the normal component that is equal to

$a_n = \dfrac{v^2}{r}= \dfrac{400}{26.5} = 15\,\text{ft/s}^2.$

The total acceleration can be obtained by means of Pythagorean theorem: $a= \sqrt{a_t^2+a_n^2} = \sqrt{6^2+15^2} = 16.2\,\text{ft/s}^2.$