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!

this is wrong

To estimate the height above the Earth's surface at 120 seconds, we can use the trapezoidal rule to approximate the distance traveled. Here are the steps to calculate the lower and upper estimates:

Step 1: Calculate the velocity differences (Δv):
Δv1 = v2 - v1 = 154 - 0 = 154 ft/s
Δv2 = v3 - v2 = 374 - 154 = 220 ft/s
Δv3 = v4 - v3 = 418 - 374 = 44 ft/s
Δv4 = v5 - v4 = 770 - 418 = 352 ft/s
Δv5 = v6 - v5 = 1320 - 770 = 550 ft/s
Δv6 = v7 - v6 = 1508 - 1320 = 188 ft/s
Δv7 = v8 - v7 = 3905 - 1508 = 2397 ft/s

Step 2: Calculate the time intervals (Δt):
Δt1 = t2 - t1 = 7 - 0 = 7 sec
Δt2 = t3 - t2 = 17 - 7 = 10 sec
Δt3 = t4 - t3 = 19 - 17 = 2 sec
Δt4 = t5 - t4 = 35 - 19 = 16 sec
Δt5 = t6 - t5 = 60 - 35 = 25 sec
Δt6 = t7 - t6 = 64 - 60 = 4 sec
Δt7 = t8 - t7 = 115 - 64 = 51 sec

Step 3: Calculate the area of each trapezoid (Δx):
Δx1 = (Δv1 + Δv2) / 2 * Δt1 = (154 + 220) / 2 * 7 = 987 ft
Δx2 = (Δv2 + Δv3) / 2 * Δt2 = (220 + 44) / 2 * 10 = 1320 ft
Δx3 = (Δv3 + Δv4) / 2 * Δt3 = (44 + 352) / 2 * 2 = 396 ft
Δx4 = (Δv4 + Δv5) / 2 * Δt4 = (352 + 550) / 2 * 16 = 7536 ft
Δx5 = (Δv5 + Δv6) / 2 * Δt5 = (550 + 188) / 2 * 25 = 10950 ft
Δx6 = (Δv6 + Δv7) / 2 * Δt6 = (188 + 2397) / 2 * 4 = 2392 ft

Step 4: Calculate the total estimated distance traveled:
Total estimated distance = Δx1 + Δx2 + Δx3 + Δx4 + Δx5 + Δx6

Lower estimate of distance traveled = Total estimated distance in feet
Upper estimate of distance traveled = Total estimated distance in feet

Now let's calculate the estimates:
Total estimated distance = 987 + 1320 + 396 + 7536 + 10950 + 2392 = 34881 ft

Lower estimate of distance traveled = 34881 ft
Upper estimate of distance traveled = 34881 ft

Reported answers accurate to 1 place:
Lower estimate of distance traveled = 34881 ft
Upper estimate of distance traveled = 34881 ft

Note: The distances are reported in feet because the velocity data is given in ft/s.

To estimate the height above the Earth's surface at 120 seconds, we can use the velocity data to estimate the distance traveled. To do this, we need to calculate the area under the velocity vs. time graph.

First, let's tabulate the given velocity data:

t (sec) v (ft/s)
0 0
7 154
17 374
19 418
35 770
60 1320
64 1508
115 3905

The intervals between the given time values are not equally spaced, so we cannot use a simple formula to find the area. Instead, we will use a method called trapezoidal approximation.

To estimate the distance traveled at 120 seconds, we need to find the areas of each trapezoid formed by adjacent data points.

The first trapezoid is formed between the given data points at t=0 and t=7. To find the area of this trapezoid, we can use the formula for the area of a trapezoid:

Area = ((v2 + v1) / 2) * (t2 - t1)

where v2 and v1 are the velocities at t2 and t1, respectively, and t2 and t1 are the time values.

Calculating the area for the first trapezoid:
Area1 = ((154 + 0)/2) * (7 - 0)
= (154/2) * 7
= 1078/2
= 539 ft

Similarly, we can calculate the areas for the other trapezoids:

Area2 = ((374 + 154)/2) * (17 - 7)
= (528/2) * 10
= 2640 ft

Area3 = ((418 + 374)/2) * (19 - 17)
= (792/2) * 2
= 792 ft

Area4 = ((770 + 418)/2) * (35 - 19)
= (1188/2) * 16
= 9504 ft

Area5 = ((1320 + 770)/2) * (60 - 35)
= (2090/2) * 25
= 26125 ft

Area6 = ((1508 + 1320)/2) * (64 - 60)
= (2828/2) * 4
= 5656 ft

Area7 = ((3905 + 1508)/2) * (115 - 64)
= (5413/2) * 51
= 138481.5 ft

Now, we can sum up all the areas to get the lower and upper estimates of the distance traveled:

Lower estimate of distance traveled = Area1 + Area2 + Area3 + Area4 + Area5 + Area6 + Area7
= 539 + 2640 + 792 + 9504 + 26125 + 5656 + 138481.5
≈ 176138.5 ft

Upper estimate of distance traveled = Area1 + Area2 + Area3 + Area4 + Area5 + Area6 + Area7 + ((1508 + 3905)/2) * (120 - 115)
= 176138.5 + ((5413/2) * 5)
≈ 179147.5 ft

Therefore, the lower estimate of distance traveled is approximately 176138.5 ft, and the upper estimate is approximately 179147.5 ft.

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.