A model rocket is fired vertically upward from rest. Its acceleration for the fi

Question

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t)=60t, at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to -18ft/sec in 5 seconds. The rocket then "floats" to the ground at that rate. (a) Determine the position function 's' and the velocity function v (for all times 't'). Sketch the graphs of 's' and 'v'. (b)At what time does the rocket reach its maximum height, and what is that height? (c)At what time does the rocket land? Please use antiderivatives to solve the problem. Thanks.

Explanation / Answer

Break the flight down into intervals: a. t = 0 to 3, a(t) = 60t (+ indicates acceleration up) v(t) = 30t^2 (initial velocity is zero)...........v(3) = 270 ft/sec s(t) = 10t^3 (initial height is zero)..............s(3) = 270 ft b t= 3 to t = 17, a(t) = -32 ft/sec^2 (downward acceleration of gravity) v(t) = - 32(t - 3) + 270.......velocity reaches zero at 11.4375 seconds; max height reached, and velocity is down. s(t) = -16(t - 3)^2 + 270(t - 3) + 270......s(11.4375) = 1409.0625 ft. v(17) = - 178 ft/sec and s(17) = 914 ft. These are the initial conditions for the next interval. c. In five seconds, velocity changes from -178 to -18 ft/sec, so the acceleration is again positive, this time + 32 ft/sec^2. t = 17 to t = 22. a = + 32 v(t) = 32(t - 17) - 178...............v(22) = - 18 ft/sec s(t) = 16(t - 17)^2 - 178(t - 17) + 914...........s(22) = 424 ft d. Rest of the flight, velocity is constant at -18 ft/sec. 424 ft/(18 ft/sec) = 23.56 sec........rocket lands at 45.56 sec.

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