Chapter 4.1, Problem 53E

(a)

To determine

To find:the distance from Earth after 30 seconds, 1 minute and 2 minutes.

Answer to Problem 53E

  $\begin{array}{l}t=30,7380\text{feet}\\ t=60,\text{ 29520 feet}\\ t=120,\text{ 118080 feet}\end{array}$

Explanation of Solution

Given information: Aeronautics: At liftoff, the space shuttle Discovery has a constant acceleration a, of 16.4 feet per second squared. The function $d\left(t\right)=\frac{1}{2}a{t}^2$ can be used to determine the distance from Earth for each time interval, t, after takeoff.

Calculation:

Distance from Earth after t seconds $=\frac{1}{2}a{t}^2$

Acceleration, $a=16.4{\text{ feet/sec}}^2,t=\text{time in second}$

  $t=30$

Distance from Earth after 30 seconds $=\frac{1}{2}16.4\ast {30}^2=7380\text{ feet}$

  $t=60$

Distance from Earth after 1 minute (60 seconds) $=\frac{1}{2}16.4\ast {60}^2=29520\text{ feet}$

  $t=120$

Distance from Earth after 2 minutes (120 seconds) $=\frac{1}{2}16.4\ast {120}^2=118080\text{ feet}$

(b)

To determine

To find:the distance from Earth change, if the time the space shuttle is in flight doubles.

Answer to Problem 53E

The distance from Earth increases 4 times

Explanation of Solution

Given information: Aeronautics: At liftoff, the space shuttle Discovery has a constant acceleration a, of 16.4 feet per second squared. The function $d\left(t\right)=\frac{1}{2}a{t}^2$ can be used to determine the distance from Earth for each time interval, t, after takeoff.

Calculation:

Distance from Earth after 30 seconds, 1 minute and 2 minutes are in order 7380, 29520, 118080 feet.

The common ratio of this sequence is 4.

  $\begin{array}{l}\frac{118080}{29520}=4\\ \frac{29520}{7380}=4\end{array}$

That is the time the space shuttle is in flight doubles, the distance from Earth increases 4 times.

That is the time the space shuttle is in flight doubles, the distance from Earth is in a geometric sequence.