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.

Expert's answer

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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