For the scotch yoke mechanism shown, the acceleration of point a is defined by the relation a = −1.08sinkt − 1.44coskt, where a

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For the scotch yoke mechanism shown, the acceleration of point a is defined by the relation a = −1.08sinkt − 1.44coskt, where a

and t are expressed in m/s2 and seconds, respectively, and k = 3 rad/s. knowing that x = 0.16 m and v = 0.36 m/s when t = 0, determine the velocity and position of point a when t =0.35 s. (round the final answers to one decimal place.)

Mathematics

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AnnZ [12.3K] · Answer 1 · 2026-03-21T07:56:38+03:00

<span>At t=0.35s, the position is 0.2 m. At the same time, the velocity is -0.2 m/s. In collegiate mathematics, the term "acceleration" implies that you are working with the second derivative of a function that describes the position of point a. To determine the velocity, you're looking for the first derivative of that function, and for the position, you're interested in the function itself. Let's compute the necessary anti-derivatives. f''(t) = -1.08 sin(kt) - 1.44 cos(kt) Taking the integral of f''(t) with respect to t yields: f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C To find C, we know that v = 0.36 m/s at t=0. Let's substitute: f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C 0.36 = (1.08 cos(3*0) - 1.44 sin(3*0))/3 + C 0.36 = (1.08 cos(0) - 1.44 sin(0))/3 + C 0.36 = (1.08*1 - 1.44*0)/3 + C 0.36 = 0.36 + C 0 = C Thus, we have f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k. Now we will find the actual function by integrating once more: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C Now let’s find out C: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C 0.16 = (1.08 sin(3*0) + 1.44 cos(3*0))/3^2 + C 0.16 = (1.08 sin(0) + 1.44 cos(0))/9 + C 0.16 = (1.08*0 + 1.44*1)/9 + C 0.16 = 1.44/9 + C 0.16 = 0.16 + C 0 = C Thus, C = 0, and the position function is expressed as: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2. Using the position and velocity functions, we can now find the desired results: Position: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 f(t) = (1.08 sin(3*0.35) + 1.44 cos(3*0.35))/3^2 f(t) = (1.08 sin(1.05) + 1.44 cos(1.05))/9 f(t) = (1.08*0.867423226 + 1.44*0.497571048)/9 f(t) = (0.936817084 + 0.716502309)/9 f(t) = 1.653319393/9 f(t) = 0.183702155. Thus, at t=0.35s, the position of point a is 0.2 m. Now calculating the velocity: f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k f'(t) = (1.08 cos(3*0.35) - 1.44 sin(3*0.35))/3 f'(t) = (1.08 cos(1.05) - 1.44 sin(1.05))/3 f'(t) = (1.08*0.497571048 - 1.44*0.867423226)/3 f'(t) = (0.537376732 - 1.249089445)/3 f'(t) = -0.711712713/3 f'(t) = -0.237237571. Therefore, the velocity at t = 0.35s is -0.2 m/s</span>