BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5.1, Problem 14E
To determine

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

Expert Solution

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=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=titi1

Here, the time of ith interval is ti and the time interval of (i1) is ti1.

Substitute 6 for n and titi1 for Δti and expand the terms in Equation (1).

d=i=16v(ti)(titi1)=[[v(t1)×(t1t0)]+[v(t2)×(t2t1)]+[v(t3)×(t3t2)]+[v(t4)×(t4t3)]+[v(t5)×(t5t4)]+[v(t6)×(t6t5)]] (2)

Here, the time velocity t1 is v(t1), the difference in time period for an interval i=1 is (t1t0), the time velocity t2 is v(t2), the difference in time period for an interval i=2 is (t2t1), the time velocity t3 is v(t3), the difference in time period for an interval i=3 is (t3t2), the time velocity t4 is v(t4), the difference in time period for an interval i=4 is (t4t3), the time velocity t5 is v(t5), the difference in time period for an interval i=5 is (t5t4), the time velocity t1 is v(t6), and the difference in time period for an interval i=6 is (t6t5).

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)×(t1t0)]+[v(t2)×(t2t1)]+[v(t3)×(t3t2)]+[v(t4)×(t4t3)]+[v(t5)×(t5t4)]+[v(t6)×(t6t5)]]=[[185×(100)]+[319×(1510)]+[447×(2015)]+[742×(3220)]+[1325×(5932)]+[1445×(6259)]]=1,850+1,595+2,235+8,904+35,775+4,335=54,694ft

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

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