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 7.5 minutes at this speed.

(a) what distance, in km, does the car travel?
(b) what is the magnitude of the car's displacement, in km, from its initial position?
(c) what is the speed of the car in m/s?

(a) 12.5 km

(b) 0 km

(c) 27.78 m/s

Explanation:

100 km/h = 100 km/h * 1000m/km * 1/60 h/minute * 1/60 minute/second = 27.78 m/s

7.5 minutes = 7.5*60 = 450 s

(a) Distance is speed times time duration

s = vt = 27.78 * 450 = 12500 m or 12.5 km

(b) As the car is moving in a circular track with circumference of 12.5 km. This means that after the car drives a distance of 12.5km, it's back to the initial point. So the total magnitude of the car displacement is 0 km

(c) 27.78 m/s

a) 12.5km

b) 0 km

c) 27.78m/s

Explanation:

Given:

Circumference of the circular track = 12.5km

Speed of the car = 100km/hr

Time = 7.5 minutes = 7.5/60 hours

a) distance is a scalar quantity ( consider magnitude only irrespective of the direction)

distance = speed × time

distance = 100km/hr × 7.5/60 hr

distance = 12.5 km

b) displacement is a vector quantity ( take into consideration magnitude and direction)

Given that the circumference of the track is 12.5 km and the car have moved a distance of 12.5km, this means the car have moved one complete cycle back to its starting point. Therefore the car would be at the starting point with zero (0) displacement.

c) given speed = 100km/hr

1 hour = 3600 seconds

1 km = 1000m

Speed (m/s) = 100km/hr ×1000m/km ÷ 3600sec/hr

Speed (m/s) = 27.777777m/s

Speed (m/s) = 27.78m/s

increase masses, decrease separation

a. spokes of a wheel.

as indicated in the question radial contains rad, also found in radius, so the fracture's would span the radius.