When estimating distances from a table of velocity data, it is not necessary that the time intervals are equally spaced. After a space ship is launched, the following velocity data is obtained. Use these data to estimate the height above the Earth's surface at 120 seconds.

Question

When estimating distances from a table of velocity data, it is not necessary that the time intervals are equally spaced. After a space ship is launched, the following velocity data is obtained. Use these data to estimate the height above the Earth's surface at 120 seconds.
t (sec)	v (ft/s)
0	0
7	154
17	374
19	418
35	770
60	1320
64	1508
115	3905

lower estimate of distance traveled =  miles

upper estimate of distance traveled =  miles

Report answers accurate to 1 places. This is not meant to be a trick question...be careful of the UNITS!

Anna · Accepted Answer

this is wrong

Steve · Answer

If v is viewed as a piecewise linear function, then the slopes of the various segments are

154/7 = 21
(374-154)/(17-7) = 22
(418-374)/(19-17) = 22
(770-418)/(35-19) = 22
(1320-770)/(60-35) = 22
(1508-1320)/(64-60) = 47
(3905-1508)/(115-64) = 47

Looks like the 2nd stage kicked in at t=60. The acceleration is
22 ft/s^2 for 60s
47 ft/s^2 from then on

So, the distance traveled at time t>60 is

s = (1/2)(22)(60^2) + 1320(t-60) + (1/2)(47)(t-60)^2 feet

You will need to convert that to miles.

Not sure about the uncertainty, since it appears that the change in acceleration was at exactly t=60. Otherwise, there would have been an interval where the acceleration was between 22 and 47.