MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK hapter 02, Problem 022. T
ID: 1524024 • Letter: M
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MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK hapter 02, Problem 022. The position of a particle moving along the x axis depends on the time according to the equation x ds. bts, where xis in meters and t in seconds. Let cand b have numerical values 2.0 m/s5 and 1.3 m/ss, respectively. From t-o.osto 1.9 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s. Units (a) Number Units (b) Number (c) Number Units (d) Number (e) Number Number (g) Number (h) Number (i) Number Click if you would like to show work for this questioni Qpen Show Work Question Attemptst o of 6 used SAvr OR LATERExplanation / Answer
Given that the position of the particle, x = c*t^5 - b*t^6
where, c = 2.0 m/s^5, b = 1.3 m/s^6
(a) Displacement of the particle from t = 0 to t = 1.9 s
x(1.9) = 2.0*(1.9)^5 - 1.3*(1.9)^6 = -11.64 m.
And x(0) = 0
So, the requisite displacement = -11.64 - 0 = -11.64 m
(b) v = dx/dt = 5*2*(t)4 - 6*1.3*(t)^5 = 10t^4 - 7.8t^5
So, v(1) = 10 - 7.8 = 2.2 m/s
(c) v(2) = 10*2^4 - 7.8*2^5 = 160 - 249.6 = -89.6 m/s
(d) v(3) = 10*3^4 - 7.8*3^5 = 810 - 1895.4 = -1085.4 m/s
(e) v(4) = 10*4^4 - 7.8*4^5 = 2560 - 7987.2 = - 5427.2 m/s
(f) a = dv/dt = 40t^3 - 39t^4
a(1) = 40-39 = 1 m/s^2
(g) a(2) = 40*2^3 - 39*2^4 = 320 - 624 = -304 m/s^2
(h) a(3) = 40*3^3 - 39*3^4 = 1080 - 3159 = - 2079 m/s^2
(i) a(4) = 40*4^3 - 39*4^4 = 2560 - 9984 = - 7424 m/s^2.
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