**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=∑i=1nv(ti)Δti (1)

Here, the time velocity ti is v(ti) and the time period is Δti.

Find the time period (Δti) using the relation:

Δti=ti−ti−1

Here, the time of *i*^{th }interval is ti and the time interval of (i−1) is ti−1.

Substitute 6 for *n* and ti−ti−1 for Δti and expand the terms in Equation (1).

d=∑i=16v(ti)(ti−ti−1)=[[v(t1)×(t1−t0)]+[v(t2)×(t2−t1)]+[v(t3)×(t3−t2)]+[v(t4)×(t4−t3)]+[v(t5)×(t5−t4)]+[v(t6)×(t6−t5)]] (2)

Here, the time velocity t1 is v(t1), the difference in time period for an interval i=1 is (t1−t0), the time velocity t2 is v(t2), the difference in time period for an interval i=2 is (t2−t1), the time velocity t3 is v(t3), the difference in time period for an interval i=3 is (t3−t2), the time velocity t4 is v(t4), the difference in time period for an interval i=4 is (t4−t3), the time velocity t5 is v(t5), the difference in time period for an interval i=5 is (t5−t4), the time velocity t1 is v(t6), and the difference in time period for an interval i=6 is (t6−t5).

Substitute 185 ft/s for v(t1), 0 for t0, 10 for t1, 319 ft/s for v(t2), 15 for t2, 447 ft/s for v(t3), 20 for t3, 742ft/s for v(t4), 32 for t4, 1325ft/s for v(t5), 59 for t5, 1445ft/s for v(t6), and 62 for t6 in Equation (2).

d=[[v(t1)×(t1−t0)]+[v(t2)×(t2−t1)]+[v(t3)×(t3−t2)]+[v(t4)×(t4−t3)]+[v(t5)×(t5−t4)]+[v(t6)×(t6−t5)]]=[[185×(10−0)]+[319×(15−10)]+[447×(20−15)]+[742×(32−20)]+[1325×(59−32)]+[1445×(62−59)]]=1,850+1,595+2,235+8,904+35,775+4,335=54,694 ft

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