Home › NCEA Level 1 › Physics › Motion in One Dimension › Runner A starts jogging at t = 0 s and Runner B …
Runner A starts jogging at t = 0 s and Runner B starts at t = 2.5 s. Both reach a point 64.0 m from the start at t = 25 s, each moving at a constant speed. If they keep those speeds, about how far from the start is each runner at t = 45 s?
ARunner A and Runner B are both about 64.0 m
BRunner A about 1.21 x 10^2 m, Runner B about 1.15 x 10^2 m
CRunner A and Runner B are both about 1.15 x 10^2 m
DRunner A about 1.15 x 10^2 m, Runner B about 1.21 x 10^2 m
Answer & Solution
Correct answer: D. Runner A about 1.15 x 10^2 m, Runner B about 1.21 x 10^2 m
1. Runner A has been moving for 25 s at t = 25 s, so v_A = 64.0 m / 25 s = 2.56 m/s.
2. Runner B has only been moving for 25 - 2.5 = 22.5 s, so v_B = 64.0 m / 22.5 s = 2.844 m/s.
3. At t = 45 s, Runner A has moved for 45 s: distance = 2.56 x 45 = 115.2 m, or 1.15 x 10^2 m.
4. At t = 45 s, Runner B has moved for 45 - 2.5 = 42.5 s: distance = 2.844 x 42.5 = 120.9 m, or 1.21 x 10^2 m.
_Source: OpenStax Physics (CC BY 4.0), Ch 2 "Motion in One Dimension", section 2.2 Speed and Velocity_
Related questions
A bus speeds up uniformly from 6 metres per second to 18 metres per second in 4 seconds. WWhat does the area under a velocity versus time graph give?A learner walks a long curved path from home to school. Which quantity depends only on theOn the jet car's velocity-time graph, a rectangular region (20 m/s over 30 s) contributes On a velocity-versus-time graph, what does the area under the curve represent?To find a jet car's instantaneous velocity at t = 25 s, the tangent line to its position-tOn a jet-powered car's position-time graph, two points are (0.50 s, 525 m) and (6.4 s, 200On a graph of position versus time, what physical quantity does the slope of the line repr