The Texas Motor Speedway is an oval racing track with curves of radius 229 m that are banked at an angle of 24.0°. The size of the track allows for race cars to achieve speeds in excess of 400 km/h. An unintended consequence of the banked curves and high speeds is that race car drivers experience excessive G forces, especially in the "vertical" direction (with respect to their body position), causing dizziness and loss of consciousness.

(a) A daring driver, pushing his car to the limit, experiences 3.57g of centripetal acceleration during one banking maneuver. What is the speed of his car during the maneuver? (Give your answer in km/h.)

km/h

(b) If the driver is seated so that during the banking maneuver his upper body is aligned with the direction normal to the banked curve, what is the magnitude of the component of the centripetal acceleration along the driver's upper body? (Give your answer in terms of g.)

Question

Harry Carlson

Physics

19 June, 10:56

The Texas Motor Speedway is an oval racing track with curves of radius 229 m that are banked at an angle of 24.0°. The size of the track allows for race cars to achieve speeds in excess of 400 km/h. An unintended consequence of the banked curves and high speeds is that race car drivers experience excessive G forces, especially in the "vertical" direction (with respect to their body position), causing dizziness and loss of consciousness.

(a) A daring driver, pushing his car to the limit, experiences 3.57g of centripetal acceleration during one banking maneuver. What is the speed of his car during the maneuver? (Give your answer in km/h.)

km/h

(b) If the driver is seated so that during the banking maneuver his upper body is aligned with the direction normal to the banked curve, what is the magnitude of the component of the centripetal acceleration along the driver's upper body? (Give your answer in terms of g.)

+1

Answers (1)

Know the Answer?

Ezekiel Bird · Answer 1

Given that the road is banked at an angle of 24.0°

a = 3.57g

but a = v^2/r = centripetal acceleration

v = square root (ar)

radius = 229m

v = square root (229 x 3.57g)

= square root (229 x 3.57 x 9.81)

= 89.55m/s x 1km/1000m x 3600s/1hr

= 322.29km/hr

b) component of centripetal acceleration = asinФ

= 3.57g x sin 24.0° = 14.24m/s^2

= 1.45g