The Nardo ring is a circular test track for cars. It has 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.

What distance, in km, does the car travel?

I’ve got this part: 38 km

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

I’ve done this part, too: 28 m/s

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

Can’t seem to figure this part out. The car goes .5 km past the starting point, but that’s the distance, not the displacement, since it’s traveling in a circle. I’ve tried to solve it using formulas to calculate the inner angle and the chord on the circle, but the answers aren’t making sense- they’re larger than .5 km. Any idea where I might be going wrong? Thanks!

### 4 Answers

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 circumferences DISTANCE

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/s SPEED

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 km DISPLACEMENT

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.

Nardo Test Track

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

Nardo Ring