   Chapter 3.9, Problem 72E

Chapter
Section
Textbook Problem

# A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a ( t ) =   60 t , 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 − 18 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). Sketch the graphs of 5 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?

To determine

a)

To determine & sketch:

The position function s and the velocity function v & sketch graphs of s and v.

Explanation

1) Concept:

Theorem: If F is an antiderivative of  f on an interval I, then the most general antiderivative of f on I is Fx+c, where c is an arbitrary constant.

2) Given:

α0t=60t,  0t3

3) Calculations:

Let, α0t=60t,  0t3

Taking the general antiderivative of α0t

v0t=30t2+C, where C  is the constant.

However, t=0, v0t=0

v00=3002+C

So, C=0 then v0t is

v0t=30t2

Now taking the general antiderivative of  v0t

s0t=30t33+D

s0t=10t3+D

However, t=0, s0t=0

s00=10(0)3+D

So, D=0

s0t=10t3

Now, find out the v0t & s0t at t=3

So, substitute t=3 in v0t & s0t we get,

v03=3032=30*9=270

s03=1033=10*27=270

For 3t17,

Therefore, the acceleration must be negative then the acceleration due to gravity is

α1t=dv1dt=-32

α1t=-32 ft/s

Taking the general antiderivative of α1t we get,

v1t=-32(t-3)+v0

v1t=-32(t-3)+270 And

Taking the general antiderivative of v1t we get,

s1t=-16t-32+270t-3+s0

s1t=-16t-32+270t-3+270

Now at t=17 substitute in above v1t & s1t we get,

v1t=-178

s1t=914

For 17<t22,

α2t=32

Taking the general antiderivative of α2t we get,

v2t=32

To determine

b)

To find:

The time and the height when the rocket reaches its maximum height.

To determine

c)

To obtain:

The time of landing of the rocket.

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