Find the velocity function.

Hey Pretty, good to see you again. a) Step 1: Find the velocity function. The acceleration is given by a = (4t - 13) ms^-2. Velocity v is the integral of acceleration with respect to time t. v = a \, dt = (4t - 13) \, dt v = 2t^2 - 13t + C We are given that the initial velocity is 18 ms^-1, which means v = 18 when t = 0. Substitute these values to find C: 18 = 2(0)^2 - 13(0) + C C = 18 So, the velocity function is: v = 2t^2 - 13t + 18 Step 2: Determine t when the particle is momentarily at rest. The particle is momentarily at rest when its velocity v = 0. 2t^2 - 13t + 18 = 0 We can solve this quadratic equation by factorization. We need two numbers that multiply to 2 × 18 = 36 and add up to -13. These numbers are -4 and -9. 2t^2 - 4t - 9t + 18 = 0 2t(t - 2) - 9(t - 2) = 0 (2t - 9)(t - 2) = 0 This gives two possible values for t: 2t - 9 = 0 2t = 9 t = (9)/(2) = 4.5 s t - 2 = 0 t = 2 s The values of t when the particle is momentarily at rest are 2 s and 4.5 s. b) Step 3: Find the position function. Position s is the integral of velocity with respect to time t. s = v \, dt = (2t^2 - 13t + 18) \, dt s = (2)/(3)t^3 - (13)/(2)t^2 + 18t + D For calculating distance covered, the constant D will cancel out, so we can set D=0. s(t) = (2)/(3)t^3 - (13)/(2)t^2 + 18t Step 4: Check for changes in direction within the interval [1, 3] s. From part (a), we know the particle is momentarily at rest at t = 2 s and t = 4.5 s. The value t = 2 s lies within the interval [1, 3] s. This means the particle changes direction at t=2 s. Therefore, we must calculate the distance covered in two separate intervals: [1, 2] s and [2, 3] s, and sum their absolute displacements. Step 5: Calculate the distance covered. Displacement from t=1 to t=2: s(2) = (2)/(3)(2)^3 - (13)/(2)(2)^2 + 18(2) = (16)/(3) - 26 + 36 = (16)/(3) + 10 = (16 + 30)/(3) = (46)/(3) m s(1) = (2)/(3)(1)^3 - (13)/(2)(1)^2 + 18(1) = (2)/(3) - (13)/(2) + 18 = (4 - 39 + 108)/(6) = (73)/(6) m Displacement s_1 = s(2) - s(1) = (46)/(3) - (73)/(6) = (92 - 73)/(6) = (19)/(6) m. Displacement from t=2 to t=3: s(3) = (2)/(3)(3)^3 - (13)/(2)(3)^2 + 18(3) = 18 - (117)/(2) + 54 = 72 - (117)/(2) = (144 - 117)/(2) = (27)/(2) m Displacement s_2 = s(3) - s(2) = (27)/(2) - (46)/(3) = (81 - 92)/(6) = -(11)/(6) m. Total distance covered is the sum of the absolute values of the displacements: Total distance = | s_1| + | s_2| = |(19)/(6)| + |-(11)/(6)| Total distance = (19)/(6) + (11)/(6) = (30)/(6) = 5 m The distance covered by the particle between t = 1 second and t = 3 seconds is 5 m. Send me the next one 📸