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a students asked her friend to help her measure her progress during a 4…

Question

a students asked her friend to help her measure her progress during a 400 meter race. friend a put a gps belt on the runner that would track her position as she moves around the track. friend b stood at the starting line and pointed a laser rangefinder at the runner to measure how far away she was at each point as she moved around the track.

  1. was student a measuring distance or displacement? what is the justification for your answer? use complete sentences to receive full credit.
  2. was student b measuring distance of displacement? what is the justification for your answer? use complete sentences to receive full credit.
  3. student a made the following table of the runner’s positions. use the image of the track to calculate the values of the measurement student b would have taken. record them below.
timestudent astudent b
15 s100 m112 m
30 s200 m
45 s300 m
60 s400 m

Explanation:

Step1: Understand distance and displacement

Distance is total path - length, displacement is straight - line distance from start to end.

Step2: Analyze student A's measurement

Student A uses GPS belt which tracks total path length of runner, so it's distance.

Step3: Analyze student B's measurement

Student B uses laser rangefinder at starting line to measure straight - line distance to runner's position, so it's displacement.

Step4: Calculate displacements for student B

  • At 15 s: The runner is 100 m along the straight part of the track. Using Pythagorean theorem for the semi - circular part at the end, if we consider the straight part and part of the semi - circle, the displacement $d=\sqrt{100^{2}+50^{2}}\approx112$ m.
  • At 30 s: The runner is at the opposite end of the track. The displacement is the diameter of the semi - circular part plus the straight part on the other side. The displacement is 100 m (straight part) + 100 m (diameter of semi - circle) = 200 m.
  • At 45 s: Similar to 15 s but on the other side of the track, displacement is $\sqrt{100^{2}+50^{2}}\approx112$ m.
  • At 60 s: The runner is back at the starting point, so displacement is 0 m.
TimeStudent AStudent B
15 s100 m112 m
30 s200 m200 m
45 s300 m112 m
60 s400 m0 m

Answer:

TimeStudent AStudent B
15 s100 m112 m
30 s200 m200 m
45 s300 m112 m
60 s400 m0 m