Adapted from IOAA 2025 (Mumbai), data-analysis problem 2, parts 1 and 2. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.
The table gives, for ten coronal mass ejections, the speed at which each left the surface of the Sun, its speed on reaching Earth and the time it took to arrive. Take the distance from the surface of the Sun to Earth as 214 solar radii, with a solar radius of km, and treat the space in between as empty.
Cloud
(km/s)
(km/s)
(h)
CME-A
804
470
74.5
CME-B
247
360
127.5
CME-C
523
396
103.5
CME-D
830
415
71.0
CME-E
665
400
104.5
CME-F
347
350
101.5
CME-G
446
375
99.5
CME-H
155
360
97.0
CME-I
1016
515
67.0
CME-J
683
410
54.0
(a) Find the average acceleration of each cloud in m s⁻². This was worth 3 marks. (b) A model says that , with km/s. Find and and their uncertainties from a suitable graph. This was worth 15 marks. (c) Tabulate the model's acceleration for each cloud, and find the root-mean-square deviation . This was worth 4 marks. (d) Two more clouds, CME-1 and CME-2, leave the Sun at 1044 km/s and 273 km/s. Taking each one's acceleration from the model as constant, predict how long each takes to reach Earth. This was worth 4 marks. (e) They actually arrived after 46.0 h and 74.5 h. The model counts as valid for a cloud if its prediction is within 20% of the observed time. Say whether it is valid for each. This was worth 2 marks.
A cloud with constant acceleration that starts at speed reaches a distance with a speed given by , and it takes the time .
The accelerations are those found in Section D1, from with the speeds in m/s and the time in seconds: −1.25, 0.246, −0.341, −1.62, −0.704, 0.008, −0.198, 0.587, −2.08 and −1.40 m s⁻².
With and , a table as in Example D.12 gives , m s⁻², and m s⁻². So m s⁻² and m s⁻².
The squared residuals add up to (m s⁻²)², so by (D.7) m s⁻² and m s⁻².
The model gives −1.40, 0.336, −0.524, −1.48, −0.966, 0.024, −0.284, 0.622, −2.06 and −1.02 m s⁻² for the ten clouds. The root-mean-square deviation, which the problem defines with 10 in the denominator, is m s⁻².
For CME-1 the model gives m s⁻². The distance is km km, so the cloud arrives at km/s, after h.
For CME-2 the model gives m s⁻², an arrival speed of 388 km/s and h.
The prediction for CME-1 is 4.8% longer than the observed 46.0 h, and the prediction for CME-2 is 68% longer than the observed 74.5 h.
AnswerThe model is m s⁻², with a root-mean-square deviation of 0.176 m s⁻². It predicts 48.2 h for CME-1, within 20% of the observed time, so it is valid there, and 125 h for CME-2, far outside 20%, so it is not valid there. The official solution, with a line drawn by hand, finds and m s⁻², within our uncertainties, and the same deviation and conclusions. Its uncertainties come from steepest and shallowest lines judged by eye, which are narrower than the scatter of the points suggests.
This problem also appears among the olympiad problems of Section D6, Least squares by hand, which teaches the ideas it uses. Equation and figure numbers refer to the course.