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A

Anonymous

Oct 30, 2024

Help with basic physics?

The Nardo Ring is a circular test track for cars with a circumference of 12.5 km. Cars travel around the track at a constant speed of 100 km/h. A car starts at the easternmost point of the ring and drives for 22.5 minutes at this speed.

  1. What distance, in kilometers, does the car travel? I calculated this part and found it to be 38 km.

  2. What is the speed of the car in meters per second? I’ve also worked this out and determined it to be 28 m/s.

  3. What is the magnitude of the car’s displacement, in kilometers, from its initial position? I’m having trouble with this part. The car travels 0.5 km past the starting point, but I understand that this is the distance traveled, not the displacement, since it’s moving in a circle. I’ve attempted to solve it using formulas to calculate the inner angle and the chord of the circle, but my answers are larger than 0.5 km. Can anyone help me identify where I might be going wrong? Thank you!

4 Answers

A
Anonymous

Jan 01, 2025

What distance, in km, does the car travel?

s = speed = 100 km/h

d = distance = to be determined

t = 22.5 min = 0.375 h

s = d/t

st = d

d = st

d = (100 km/h)(0.375 h) = 37.5 km = 37.5 km / 12.5 km/cir = 3 circumferencesDISTANCE

What is the speed of the car in m/s?

s = 100 km/h

100 km x 1,000 m/km = 100,000 m

1 h x 3,600 s/h = 3,600 s

s = 100,000 m / 3,600 s = 27.8 m/sSPEED

What is the magnitude of the car’s displacement, in km, from its initial position?

Since the car has traveled EXACTLY 3 circumferences, it ends up at the SAME place

where it began its journey around the track. Therefore, its displacement is ZERO.

d = displacement

d = 0 kmDISPLACEMENT

Any idea where I might be going wrong?

You went wrong by rounding off too soon. The calculated distance that the car traveled is 37.5 km, which, as it turns out, is EXACTLY three times the circumference of 12.5 km. The 0.5 km that you’re worrying so much about is actually an artifact of your unwarranted rounding off.

A
Anonymous

Jan 27, 2025

distance traveled along arc = 100*(22.5/60) = 37.5 km

theta = 37.5/12.5 = 3 rad

magnitude of displacement = 2*12.5*sin(theta/2) = 24.93 or 25 km

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