The following graph displays the velocity vs time graph for a parachutist jumping out of an aeroplane at t = 0. At what time(s) does the parachutist reach terminal velocity?

Question

BlackTom AI · Accepted Answer

To tackle this question, I’ll examine what terminal velocity means in the context of a velocity–time graph: it is the steady, constant velocity that occurs when drag balances weight, so the slope (acceleration) is zero.
Option a: 'At 27 minutes' — first, note the graph’s time axis is in seconds (0 to 60 s). The label 'minutes' is inconsistent with the graph. Even if we interpreted it as 27 s, at around 27 s there is a dramatic change (the parachute opens), which causes velocity to drop sharply from about 60 m/s to roughly 10 m/s. This is not a moment where velocity remains constant; it is a transient change, so this option is not describing terminal velocity.
Option b: 'At 55 minutes' — again, the time unit is wrong and well beyond the displayed range. The graph shows only up to 60 s, and long before 55 minutes would be reached the velocity has already changed multiple times (including the parachute opening). Therefore this choice is not consistent with the graph.
Option c: 'At 20 and 27 minutes' — converting minutes to seconds (if we treated minutes as a unit here) gives 1200 s and 1620 s, both far outside the graph’s 0–60 s range. Even if the option intended to express times within the visible 0–60 s window, the phrase '20 and 27' aligns with 20 s and 27 s, which does match key events on the graph: the velocity climbs and plateaus near 60 m/s around 20–25 s (a quasi-terminal phase before parachute deployment), then at about 27 s there is a sudden velocity change due to parachute opening, after which a new steady velocity (~10 m/s) is reached. This option attempts to include both the pre-parachute terminal-like plateau and the post-deployment terminal velocity as distinct times, but the wording is ambiguous. As a result, it is the closest to describing the two actual constant-velocity phases visible on the graph, though it’s not perfectly precise in units.
Option d: 'At 20 minutes' — if interpreted as 20 s, this corresponds to the onset of the near-constant velocity phase before the parachute opens, but the graph shows the velocity is still rising slightly and has not yet settled into the high-velocity plateau exactly at 20 s; the true steady state is approached a little after 20 s. If interpreted as 20 s, this could be considered roughly aligned with the early plateau, but it omits the later post-deployment terminal velocity, so it is incomplete.
In summary, the graph contains two distinct constant-velocity regimes: a high-velocity plateau around roughly 20–25 s before parachute deployment, and a much lower constant velocity around 27 s onward after the parachute opens. The options, however, use inconsistent time units (minutes) and do not clearly label the two regimes. The most defensible interpretation among the choices, given the graph’s events, is that there are two terminal-velocity phases observed at approximately the times around 20–25 s (pre-deployment) and just after 27 s (post-deployment).

The following graph displays the velocity vs time graph for a parachutist jumping out of an aeroplane at t = 0. At what time(s) does the parachutist reach terminal velocity?

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