This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Okay, Abbie, let's tackle this motion problem. I'll assume the position function is s(t) = t^3 - 9t^2 + 24t, as t^2 - 9t^2 + 24t would simplify to -8t^2 + 24t, which has a constant acceleration (meaning acceleration is never zero for t>0). Given the position function: s(t) = t^3 - 9t^2 + 24t 1. Find the positions when the body is momentarily at rest. Step 1: Find the velocity function v(t) by differentiating s(t) with respect to t. v(t) = (ds)/(dt) = (d)/(dt)(t^3 - 9t^2 + 24t) v(t) = 3t^2 - 18t + 24 Step 2: Set the velocity v(t) to zero to find the times when the body is momentarily at rest. 3t^2 - 18t + 24 = 0 Divide the entire equation by 3: t^2 - 6t + 8 = 0 Factor the quadratic equation: (t-2)(t-4) = 0 This gives two times when the body is at rest: t = 2 or t = 4 Step 3: Substitute these times back into the position function s(t) to find the positions. For t=2: s(2) = (2)^3 - 9(2)^2 + 24(2) s(2) = 8 - 9(4) + 48 s(2) = 8 - 36 + 48 s(2) = 20 For t=4: s(4) = (4)^3 - 9(4)^2 + 24(4) s(4) = 64 - 9(16) + 96 s(4) = 64 - 144 + 96 s(4) = 16 The positions when the body is momentarily at rest are 20 units and 16 units. 2. Find the position when its acceleration is zero. Step 4: Find the acceleration function a(t) by differentiating the velocity function v(t) with respect to t. a(t) = (dv)/(dt) = (d)/(dt)(3t^2 - 18t + 24) a(t) = 6t - 18 Step 5: Set the acceleration a(t) to zero to find the time when the acceleration is zero. 6t - 18 = 0 6t = 18 t = 3 Step 6: Substitute this time back into the position function s(t) to find the position. For t=3: s(3) = (3)^3 - 9(3)^2 + 24(3) s(3) = 27 - 9(9) + 72 s(3) = 27 - 81 + 72 s(3) = 18 The position when the acceleration is zero is 18 units. Last free one today — make it count tomorrow, or type /upgrade for unlimited.