Chapter 5.1, Problem 14E

To determine

The distance traveled by the Endeavor above the Earth’s surface after liftoff of 62 sec.

Answer to Problem 14E

The distance traveled by the Endeavor above the Earth’s surface after the liftoff of 62 sec is 54,694 ft.

Explanation of Solution

Given information:

The number of intervals till the liftoff of 62 sec is $n=6$.

The expression to find the distance traveled by the Endeavor after 62 sec of liftoff (d) using the upper estimate as shown below:

$d={\displaystyle \sum _{i=1}^{n}v\left(t_i\right)\Delta t_i}$ (1)

Here, the time velocity $t_i$ is $v\left(t_i\right)$ and the time period is $\Delta t_i$.

Find the time period $\left(\Delta t_i\right)$ using the relation:

$\Delta t_i=t_i-t_{i-1}$

Here, the time of i^th interval is $t_i$ and the time interval of $\left(i-1\right)$ is $t_{i-1}$.

Substitute 6 for n and $t_i-t_{i-1}$ for $\Delta t_i$ and expand the terms in Equation (1).

$\begin{array}{c}d={\displaystyle \sum _{i=1}^{6}v\left(t_i\right)\left(t_i-t_{i-1}\right)}\\ =\left[\begin{array}{l}\left[v\left(t_1\right)\times \left(t_1-t_0\right)\right]+\left[v\left(t_2\right)\times \left(t_2-t_1\right)\right]+\left[v\left(t_3\right)\times \left(t_3-t_2\right)\right]\\ +\left[v\left(t_4\right)\times \left(t_4-t_3\right)\right]+\left[v\left(t_5\right)\times \left(t_5-t_4\right)\right]+\left[v\left(t_6\right)\times \left(t_6-t_5\right)\right]\end{array}\right]\end{array}$ (2)

Here, the time velocity $t_1$ is $v\left(t_1\right)$, the difference in time period for an interval $i=1$ is $\left(t_1-t_0\right)$, the time velocity $t_2$ is $v\left(t_2\right)$, the difference in time period for an interval $i=2$ is $\left(t_2-t_1\right)$, the time velocity $t_3$ is $v\left(t_3\right)$, the difference in time period for an interval $i=3$ is $\left(t_3-t_2\right)$, the time velocity $t_4$ is $v\left(t_4\right)$, the difference in time period for an interval $i=4$ is $\left(t_4-t_3\right)$, the time velocity $t_5$ is $v\left(t_5\right)$, the difference in time period for an interval $i=5$ is $\left(t_5-t_4\right)$, the time velocity $t_1$ is $v\left(t_6\right)$, and the difference in time period for an interval $i=6$ is $\left(t_6-t_5\right)$.

Substitute 185 ft/s for $v\left(t_1\right)$, 0 for $t_0$, 10 for $t_1$, 319 ft/s for $v\left(t_2\right)$, 15 for $t_2$, 447 ft/s for $v\left(t_3\right)$, 20 for $t_3$, 742ft/s for $v\left(t_4\right)$, 32 for $t_4$, 1325ft/s for $v\left(t_5\right)$, 59 for $t_5$, 1445ft/s for $v\left(t_6\right)$, and 62 for $t_6$ in Equation (2).

$\begin{array}{c}d=\left[\begin{array}{l}\left[v\left(t_1\right)\times \left(t_1-t_0\right)\right]+\left[v\left(t_2\right)\times \left(t_2-t_1\right)\right]+\left[v\left(t_3\right)\times \left(t_3-t_2\right)\right]\\ +\left[v\left(t_4\right)\times \left(t_4-t_3\right)\right]+\left[v\left(t_5\right)\times \left(t_5-t_4\right)\right]+\left[v\left(t_6\right)\times \left(t_6-t_5\right)\right]\end{array}\right]\\ =\left[\begin{array}{c}\left[185\times \left(10-0\right)\right]+\left[319\times \left(15-10\right)\right]+\left[447\times \left(20-15\right)\right]\\ +\left[742\times \left(32-20\right)\right]+\left[1325\times \left(59-32\right)\right]+\left[1445\times \left(62-59\right)\right]\end{array}\right]\\ =1,850+1,595+2,235+8,904+35,775+4,335\\ =54,694\text{ft}\end{array}$

Hence, the distance traveled by the Endeavor above the Earth’s surface after the liftoff of 62 sec is 54,694 ft.