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

Question #128860
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.
1
2020-08-27T10:25:37-0400

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."

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