Given the velocity function V = 4t2 - 3t:

Step 1: Determine the time when the particle stops. The particle stops when its velocity V(t) is equal to zero. Given the velocity function V = 4t^2 - 3t: 4t^2 - 3t = 0 Factor out t: t(4t - 3) = 0 This gives two possible times: t = 0 or 4t - 3 = 0 Solving for the second time: 4t = 3 t = (3)/(4) The particle starts at t=0 and stops at t=(3)/(4). Step 2: Find the displacement function s(t). The displacement s(t) is the integral of the velocity function V(t) with respect to time t. s(t) = V(t)\,dt s(t) = (4t^2 - 3t)\,dt s(t) = 4t^2+12+1 - 3t^1+11+1 + C s(t) = (4t^3)/(3) - (3t^2)/(2) + C Assuming the particle starts at position s=0 at t=0, we have s(0) = 0, which means C=0. So, the displacement function is: s(t) = (4t^3)/(3) - (3t^2)/(2) Step 3: Calculate the displacement at the time the particle stops. The particle stops at t = (3)/(4). Substitute this value into the displacement function: s((3)/(4)) = (4)/(3)((3)/(4))^3 - (3)/(2)((3)/(4))^2 s((3)/(4)) = (4)/(3)((27)/(64)) - (3)/(2)((9)/(16)) s((1)/(4)) = (108)/(192) - (27)/(32) Simplify the first fraction: (108)/(192) = (9 × 12)/(16 × 12) = (9)/(16). s((3)/(4)) = (9)/(16) - (27)/(32) To subtract these fractions, find a common denominator, which is 32: s((3)/(4)) = (9 × 2)/(16 × 2) - (27)/(32) s((3)/(4)) = (18)/(32) - (27)/(32) s((3)/(4)) = -(9)/(32) Step 4: Determine the maximum distance travelled. The velocity function V(t) = t(4t-3) is negative for 0 < t < (3)/(4) (e.g., at t=0.5, V(0.5) = 0.5(2-3) = -0.5). This means the particle moves in one direction (the negative direction) from t=0 until it stops at t=(3)/(4). Therefore, the total distance travelled is the absolute value of the displacement. Distance = |s((3)/(4)) - s(0)| Distance = |-(9)/(32) - 0| Distance = |-(9)/(32)| Distance = (9)/(32) The maximum distance travelled by the particle before stopping is (9)/(32). Send me the next one 📸