A woman is sliding down an incline with a const. acceleration of a(0) = 2.3 m/s

Question

A woman is sliding down an incline with a const. acceleration of a(0) = 2.3 m/s2 relative to the incline. At the same time the incline is accelerating to the right at a(s) = 1.2 m/s2 relative to the ground. Letting = 34° and L = 4 m and assuming that both the woman and the incline start from rest, determine the horizontal distance traveled by the woman with respect to the ground when she reaches the bottom of the slide.

The acceleration down the triange with length L and angle theta, down and to the left. The incline is moving to the right, horizontally.

Explanation / Answer

horizontal component of acceleration of woman relative to the incline = 2.3cos(34) to left acceleration of the incline = 1.2 to the right horizontal acceleration of woman relative to ground = 2.3cos(34) - 1.2 = .71 vertical component relative to ground = 2.3sin(34) - 0 (incline does not have any acceleration in vertical direction)= 1.2861 height of the incline = Lsin(34) = 2.2368 m 2.2368 = 0 + .5*1.2861*t^2 t^2 = 3.4784 in horizontal direction S = .5*.71*t^2 = 1.2348 m the horizontal distance traveled by the woman with respect to the ground when she reaches the bottom of the slide =1.2348 m

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