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.
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Okay, Mazel, let's re-solve these problems, ensuring all the formatting is just right. Question 16: A particle moves in a straight line so that the velocity is given by V = (1)/(2)t^2 - 3t + 7, where t is the time in seconds. a) The velocity after 8 seconds Step 1: Substitute t=8 into the velocity equation. V = (1)/(2)(8)^2 - 3(8) + 7 Step 2: Calculate the value. V = (1)/(2)(64) - 24 + 7 V = 32 - 24 + 7 V = 15 The velocity after 8 seconds is 15 units/s. b) The acceleration when t=0 Step 1: Find the acceleration function by differentiating the velocity function with respect to t. a = (dV)/(dt) = (d)/(dt)((1)/(2)t^2 - 3t + 7) a = t - 3 Step 2: Substitute t=0 into the acceleration function. a = 0 - 3 a = -3 The acceleration when t=0 is -3 units/s^2. c) The minimum velocity attained Step 1: To find the minimum velocity, set the acceleration (first derivative of velocity) to zero. a = t - 3 = 0 t = 3 Step 2: Substitute t=3 into the original velocity equation to find the minimum velocity. V_min = (1)/(2)(3)^2 - 3(3) + 7 V_min = (1)/(2)(9) - 9 + 7 V_min = 4.5 - 9 + 7 V_min = 2.5 The minimum velocity attained is 2.5 units/s. Question 17: Given the graph y = (x^2 - 9)(x + 3). a) The x-intercept of the curve Step 1: Set y=0 to find the x-intercepts. 0 = (x^2 - 9)(x + 3) Step 2: Factor the term (x^2 - 9) as a difference of squares. 0 = (x-3)(x+3)(x+3) 0 = (x-3)(x+3)^2 Step 3: Solve for x. x-3 = 0 x = 3 x+3 = 0 x = -3 The x-intercepts are (3,0) and (-3,0). b) The y-intercept of the curve Step 1: Set x=0 to find the y-intercept. y = (0^2 - 9)(0 + 3) y = (-9)(3) y = -27 The y-intercept is (0,-27). c) The stationary point of the curve Step 1: Expand the equation for y. y = (x^2 - 9)(x + 3) y = x^3 + 3x^2 - 9x - 27 Step 2: Differentiate y with respect to x to find the gradient function. (dy)/(dx) = 3x^2 + 6x - 9 Step 3: Set (dy)/(dx) = 0 to find the x-coordinates of the stationary points. 3x^2 + 6x - 9 = 0 x^2 + 2x - 3 = 0 (x+3)(x-1) = 0 x = -3 or x = 1 Step 4: Substitute these x-values back into the original curve equation to find the corresponding y-coordinates. For x=-3: y = (-3)^3 + 3(-3)^2 - 9(-3) - 27 = -27 + 27 + 27 - 27 = 0 For x=1: y = (1)^3 + 3(1)^2 - 9(1) - 27 = 1 + 3 - 9 - 27 = -32 The stationary points are (-3,0) and (1,-32). d) The nature of the stationary point in c above Step 1: Find the second derivative of the curve. (d^2y)/(dx^2) = (d)/(dx)(3x^2 + 6x - 9) (d^2y)/(dx^2) = 6x + 6 Step 2: Evaluate the second derivative at each stationary point. For x=-3: (d^2y)/(dx^2)|_x=-3 = 6(-3) + 6 = -18 + 6 = -12 Since (d^2y)/(dx^2) < 0, the point (-3,0) is a maximum point. For x=1: (d^2y)/(dx^2)|_x=1 = 6(1) + 6 = 6 + 6 = 12 Since (d^2y)/(dx^2) > 0, the point (1,-32) is a minimum point. So, (-3,0) is a maximum point while (1,-32) is a minimum point. e) Sketch the curve. The sketch should show: • X-intercepts at (3,0) and (-3,0). • Y-intercept at (0,-27). • A maximum point at (-3,0). • A minimum point at (1,-32). The curve rises to a maximum at (-3,0), then decreases through (0,-27) to a minimum at (1,-32), and then increases, passing through (3,0). The provided image shows a correct sketch of the curve. Got more? Send 'em!