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) = 66t, at which time the fuel is exhausted and it becomes a freely "falling" body.Fifteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to

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ft/s 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).
v(t) =

33t2+v0

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s(t) =

Sketch the graph of v.

Sketch the graph of s.

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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