This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the problems. 2.3 Speedy Gonzales runs from point A in a straight line, coming to rest at point B. The distance in a straight line from A is given by the formula s = 3t^3 - (1)/(2)t^4 where t is measured in seconds and s in metres. 2.3.1 An expression for the velocity of Speedy Gonzales in terms of t Velocity v is the derivative of distance s with respect to time t. Step 1: Differentiate the given distance function s with respect to t. v = (ds)/(dt) = (d)/(dt)(3t^3 - (1)/(2)t^4) v = 3(3t^3-1) - (1)/(2)(4t^4-1) v = 9t^2 - 2t^3 The expression for velocity is v = 9t^2 - 2t^3. 2.3.2 The time taken to reach B Speedy Gonzales comes to rest at point B, which means its velocity v is 0. Step 1: Set the velocity expression from 2.3.1 to zero. 9t^2 - 2t^3 = 0 Step 2: Factor out the common term t^2. t^2(9 - 2t) = 0 Step 3: Solve for t. This gives two possible solutions: t^2 = 0 or 9 - 2t = 0. t=0 represents the starting point A. For 9 - 2t = 0: 2t = 9 t = (9)/(2) = 4.5 seconds The time taken to reach B is 4.5 seconds. **2.