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.
(0,0) (5,110) (17, 374) (25,550) (44,968) (60,1320) (69,1743) (129,4563) are the data points
questions
what is the lower estimate of distance traveled
what is the upper estimate of distance traveled
If you plot the points, it appears to be piecewise linear, with a slope change at x=60
So, interpolating between 69 and 129, we get
1743 + (51/60)(4563-1743) = 4140
To estimate the distance traveled, we can approximate it using the trapezoidal rule, which involves summing up the areas of trapezoids formed by adjacent data points.
First, let's calculate the height at 120 seconds using the given velocity data.
To estimate the height, we need to determine the velocity at 120 seconds. Since the time intervals are not evenly spaced, we need to interpolate the velocity at 120 seconds using the given data points.
For the given data points (t, v), we can interpolate the velocity at 120 seconds using linear interpolation. Linear interpolation can be done using the formula:
v = v1 + (v2 - v1) * ((t - t1) / (t2 - t1))
where (t1, v1) and (t2, v2) are the data points that form the range in which our target time, 120 seconds, lies.
From the given data, we can identify the range that contains 120 seconds:
(69, 1743) and (129, 4563)
Using the linear interpolation formula, we can calculate the velocity at 120 seconds:
v120 = 1743 + (4563 - 1743) * ((120 - 69) / (129 - 69))
= 1743 + 2820 * (51 / 60)
= 1743 + 2427
v120 ≈ 4170
Now that we have the interpolated velocity at 120 seconds, we can estimate the distance traveled.
Since the time intervals are not equal, we'll use the trapezoidal rule to approximate the distance traveled between each data point.
To estimate the lower and upper bounds of the distance traveled, we'll calculate the sum of the areas of the trapezoids formed by each adjacent pair of data points. This sum will give us an approximation of the distance.
Let's calculate the lower estimate of the distance traveled:
Lower estimate = Sum of (0.5 * (v2 + v1) * (t2 - t1)) for each adjacent data point pair
Using the given data:
Lower estimate = (0.5 * (110 + 0) * (5 - 0)) + (0.5 * (374 + 110) * (17 - 5)) + (0.5 * (550 + 374) * (25 - 17)) + (0.5 * (968 + 550) * (44 - 25)) + (0.5 * (1320 + 968) * (60 - 44)) + (0.5 * (1743 + 1320) * (69 - 60)) + (0.5 * (4170 + 1743) * (120 - 69)) + (0.5 * (4563 + 4170) * (120 - 129))
Lower estimate = 220 + 6157 + 8495 + 22819 + 49312 + 64801.5 + 108837.5 - 24562.5
Lower estimate ≈ 265180
Now, let's calculate the upper estimate of the distance traveled:
Upper estimate = Sum of (0.5 * (v1 + v2) * (t2 - t1)) for each adjacent data point pair
Using the given data:
Upper estimate = (0.5 * (0 + 110) * (5 - 0)) + (0.5 * (110 + 374) * (17 - 5)) + (0.5 * (374 + 550) * (25 - 17)) + (0.5 * (550 + 968) * (44 - 25)) + (0.5 * (968 + 1320) * (60 - 44)) + (0.5 * (1320 + 1743) * (69 - 60)) + (0.5 * (1743 + 4563) * (120 - 69)) + (0.5 * (4563 + 4170) * (120 - 129))
Upper estimate = 275 + 2420 + 5610 + 20960 + 41008 + 57092.5 + 223417.5 - 22804.5
Upper estimate ≈ 328879.5
Therefore, the lower estimate of the distance traveled is approximately 265180 and the upper estimate is approximately 328879.5 units.