Question

0/1 points I Provious Answers SCalcET6 5.1.014.MI. Ask Your Teacher 11. My Notes When we estimate distances from velocity data, it is sometimes necessary to use times to, ti, t2, t3,. . . that are not equally spaced. We can still estimate distances using the time periods Attt 1. For example, a space shuttle was launched on a mission, the purpose of which was to install a new motor between liftoff and the jettisoning of the solid rocket boosters. Use these data to estimate the height, h, above Earth's surface a satellite. The table provideed gives the velocity data for the shuttle f the space shuttle, 62 seconds after liftoff. (Give the upper from the data.) approximation ava Event Time (s) Velocity (ft/s) Launch 0 Begin roll maneuver 10 180 End roll maneuver 15 319 Throttle to 89% 20 453 Throttle to 67% 32 742 Throttle to 104% 59 1100 Maximum dynamic pressure 62 1430 Solid rocket booster separation 4151 125 Need Help? Master ItTalk to a Tutor Save Progress Submit Answer

Answer 1

Distance covered for first 10 seconds = 180*10=1800 ft

Distance covered for 10-15 second     = 319*5 = 1595 ft

Distance covered for 15-20 second     = 453*5 = 2265 ft

Distance covered for 20-32 second    = 742*12 = 8904 ft

Distance covered for 32-59 second   = 1100*27 = 29700 ft

Distance covered for 59-62 second   = 1430*3 = 4290 ft

so total distance upto 62 seconds = 1800+1595+2265+8904+29700+4290=48554 ft