find the position function s(t)describing the motion of the object.
Calculus

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base of a solid in the xyplane is the 1st quadrant region bounded y=x&y=x^2cross section of solid perpendicular to xaxis are equilateral triangles. whats the volume,in cubic units of solid? Determine if Mean Value Theorem for integrals applies to the function f(x)=x^39x on the interval [1,1]if so find xcoordinates of points guaranteed to exist by theorem.object has a constant acceleration of 40 ft/sec^2,an inital velocity of 20ft/sec,&inital posit of10ft find function s(t)descrbe motion.
The graphs y = x and y = x^{2} intersect at the points (0, 0) and (1, 1). So, V = ∫_{0}^{1} A(x)dx, where A(x) is the area of the crosssection of the solid by a plane perpendicular to the xaxis. Since the area of an equilateral triangle with the side a equals A = (√(3)/4)a^{2}, the function A(x) = (√(3)/4)(x – x^{2})^{2} and the volume V = ∫_{0}^{1} (√(3)/4)(x – x^{2})^{2}dx = (√(3)/4)∫_{0}^{1}(x^{2} – 2x^{3 }+ x^{4})dx =(√(3)/4)(x^{3}/3 – 2x^{4}/4 + x^{5}/5)]_{0}^{1 }= (√(3)/4)(1/3 – 1/2 + 1/5) = √(3)/120.
According to the Mean Value Theorem ∫_{–1}^{1}(x^{3} – 9x) dx = 2f(x*), where x* lies between –1 and 1. However, the integral equals 0 (the integrand is an odd function and the interval of integration is symmetric with respect to 0), thus, f(x*) = 0. The only possible value of x* is 0 (x^{3} – 9x = 0 has also solutions ±3 but they do not lie between –1 and 1).
The law of motion can be described as s(t) = s_{0} + v_{0}t + at^{2}/2 = 10 – 20t + 20t^{2}, where time t is expressed in seconds and the position s in feet.
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