Suppose that a model rocket blasts off, straight up, at time t = 0, with a vertical displacement given by a function f(t) for t0. During the first two seconds of the launch (0 t 2), the engine is burning, and the rocket is accelerating upward with a constant second derivative f"(t) = 40 (metres/sec). At t = 2, the engine stops burning, and gravity takes over, such that the rocket is then accelerating downward with a constant second derivative of f"(t) = -10 (metres/sec) for all t > 2. (i) Assume that The rocket begins (at time t =0) with a vertical displacement of zero (f(0) = 0). The rocket begins (at time t = 0) with a vertical velocity of zero (f'(0) = 0). The vertical displacement f(t) is continuous for all t > 0. The vertical displacement f(t) is differentiable for all t > 0. Solve for the vertical displacement f(t) of the rocket for all t2 0. (NOTE: Your solution should be a piecewise function of t, and you do not need to worry about what happens to the rocket after it hits the ground). (ii) What is the maximum height achieved by the rocket (in metres)? Justify your answer. (iii) At what time (in seconds) does the rocket hit the ground? Approximate your answer to two decimal places. Let f(x)=x and for each real number n 21, define the function 9n (2) = 1 - 1 -|x| n (i) Find the area An bounded between f and gn. Your answer should depend on n. (ii) Find the area A bounded between f and the constant function c(x) = 1. (iii) Notice that as n becomes very large, A,, approaches A. Explain why this is true by referring to the relationship between gn and c.