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.)

Position at t=0.35s is 0.2 m Velocity at t = 0.35s is -0.2 m/s Since this is college level mathematics, the use of the word "acceleration" should indicate to you that you've been given the 2nd derivative of a function specifying the location of point a. And since you've been asked for the velocity, you know that you want the 1st derivative of the function. And since you've also been asked for the position, you also want the function itself. So let's calculate the desired anti-derivatives. f''(t) = -1.08 sin(kt) - 1.44 cos(kt) The integral of f''(t) with respect to t is: f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C In order to find out what C is, we know that at time t=0, v = 0.36 m/s. So let's plug in the values and see what C is: 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 So the first derivative will be f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k Now to get the actual function by integrating again. Giving: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C And let's determine what C is: 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 So C = 0 and the position function is: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 So now, we can use out position and velocity functions to get the desired 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 So the position of point a at t=0.35s is 0.2 m Now for 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 So the velocity at t = 0.35s is -0.2 m/s