A particle moves in a straight line. Its velocity, v m/s, at time t seconds is given by v = 3t² - 10t + 8. Find the distance traveled by the particle between t = 1 second and t = 3 seconds.

Hey ~Gracina~, good to see you again. It looks like you're asking me to continue with the previous problem. Let's pick up where we left off. Problem: A particle moves in a straight line. Its velocity, v m/s, at time t seconds is given by v = 3t² - 10t + 8. Find the distance traveled by the particle between t = 1 second and t = 3 seconds. Solution (continued): Step 3: Calculate the position of the particle at the start, end, and critical points using s(t) = t^3 - 5t^2 + 8t. We had: s(1) = 1^3 - 5(1)^2 + 8(1) = 1 - 5 + 8 = 4 m Continuing with s(4/3): s(4/3) = ((4)/(3))^3 - 5((4)/(3))^2 + 8((4)/(3)) s(4/3) = (64)/(27) - 5((16)/(9)) + (32)/(3) s(4/3) = (64)/(27) - (80)/(9) + (32)/(3) To combine these, find a common denominator, which is 27: s(4/3) = (64)/(27) - (80 × 3)/(9 × 3) + (32 × 9)/(3 × 9) s(4/3) = (64)/(27) - (240)/(27) + (288)/(27) s(4/3) = (64 - 240 + 288)/(27) = (112)/(27) m Now, calculate s(2) and s(3): s(2) = 2^3 - 5(2)^2 + 8(2) = 8 - 5(4) + 16 = 8 - 20 + 16 = 4 m s(3) = 3^3 - 5(3)^2 + 8(3) = 27 - 5(9) + 24 = 27 - 45 + 24 = 6 m Step 4: Calculate the absolute displacements for each interval and sum them to find the total distance. The critical points are t = 4/3 and t = 2. The intervals are [1, 4/3], [4/3, 2], and [2, 3]. Displacement for [1, 4/3]: |s(4/3) - s(1)| = |(112)/(27) - 4| = |(112)/(27) - (108)/(27)| = |(4)/(27)| = (4)/(27) m Displacement for [4/3, 2]: |s(2) - s(4/3)| = |4 - (112)/(27)| = |(108)/(27) - (112)/(27)| = |-(4)/(27)| = (4)/(27) m Displacement for [2, 3]: |s(3) - s(2)| = |6 - 4| = |2| = 2 m Total distance traveled: Distance = (4)/(27) + (4)/(27) + 2 Distance = (8)/(27) + 2 Distance = (8)/(27) + (54)/(27) Distance = (62)/(27) m The total distance traveled by the particle between t = 1 second and t = 3 seconds is (62)/(27) m. 3 done, 2 left today. You're making progress.