Which equation gives the altitude a, in meters, of the balloon t minutes after it is released?

Question

During a weather balloon launch, the balloon rises at a constant rate after it is released from a platform above ground level. Technicians record that 12 minutes after release the balloon is at an altitude of 4,100 meters, and 20 minutes after release it is at an altitude of 6,500 meters.

Accepted Answer

B. a = 300t + 500. The constant rate is (6,500 - 4,100)/(20 - 12) = 2,400/8 = 300 meters per minute. Using (12, 4,100): 4,100 = 300(12) + b, so b = 4,100 - 3,600 = 500, giving a = 300t + 500. Choice A assumes the balloon starts at ground level (altitude 0 at t = 0), ignoring the platform. Choice C mistakes the recorded altitude 4,100 for the initial altitude. Choice D uses the total change in altitude, 2,400 meters, as the per-minute rate instead of dividing by the 8-minute change in time.

Answer

A. a = 300t — Your rate of 300 m/min is right, but the balloon is released from a platform above the ground, so the altitude at t = 0 isn't 0 — it's 500 meters.

Answer

C. a = 300t + 4,100 — 4,100 meters is the altitude at t = 12 minutes, not at t = 0. Substituting (12, 4,100) into a = 300t + b gives a starting altitude of 500 meters.

Answer

D. a = 2,400t + 500 — 2,400 is the total change in altitude over 8 minutes, not the per-minute rate. Divide: 2,400/8 = 300 meters per minute, which gives a = 300t + 500.

Algebra practice question

Answer: B. a = 300t + 500

Why the other answers are wrong

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