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 3.62g of centripetal acceleration during one banking maneuver. What is the speed of his car during the maneuver? (Give your answer in 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.)
(a) Well, if this daring driver is experiencing 3.62g of centripetal acceleration, then we can calculate their speed during the maneuver using a bit of physics. The centripetal acceleration is given by the equation:
a = v^2 / r
Where a is the centripetal acceleration, v is the speed, and r is the radius of the curve. We can rearrange the equation to solve for the speed:
v = sqrt(a * r)
Plugging in the given values, we have:
v = sqrt(3.62g * 229 m)
But since we want the answer in km/h, we need to convert meters to kilometers and seconds to hours. Conveniently, 1 m/s = 3.6 km/h:
v = sqrt(3.62 * 9.8 m/s^2 * 229 m) * 3.6 km/h
After doing the math, we find that the speed of the car during the maneuver is approximately 250 km/h.
(b) Now, if the driver's upper body is aligned with the direction normal to the banked curve, we need to calculate the magnitude of the component of the centripetal acceleration along the driver's upper body. This component will be the vertical component since the upper body is perpendicular to the horizontal curve.
To find this, we need to calculate the vertical component of the centripetal acceleration using:
a_vertical = a * sin(θ)
Where θ is the angle of the banked curve. Since the angle is given as 24.0°, we can plug in the values:
a_vertical = 3.62g * sin(24.0°)
Simplifying, we find that the magnitude of the component of the centripetal acceleration along the driver's upper body is approximately 1.50g. So, the driver will feel about 1.50 times their own weight pressing down on their upper body during the maneuver. Quite a weighty experience, wouldn't you say?
(a) To find the speed of the car during the maneuver, we need to apply the centripetal acceleration formula:
a = v² / r
Where:
a = centripetal acceleration (in m/s²)
v = velocity of the car (in m/s)
r = radius of the curve (in m)
We are given the centripetal acceleration a = 3.62g and the radius r = 229 m. Converting the acceleration to m/s²:
a = 3.62 * 9.8 m/s² = 35.436 m/s²
Now we can rearrange the formula and solve for v:
v² = a * r
v² = 35.436 m/s² * 229 m
v² = 8117.784 m²/s²
Taking the square root of both sides:
v = √(8117.784 m²/s²) ≈ 90.07 m/s
Converting the velocity to km/h:
v = 90.07 m/s * (3600 s/1 hr) * (1 km/1000 m) ≈ 324.25 km/h
Therefore, the speed of the car during the maneuver is approximately 324.25 km/h.
(b) The component of the centripetal acceleration along the driver's upper body can be calculated using trigonometry. The angle of the banked curve is given as 24.0°. Using this angle, we can find the component of the centripetal acceleration in the vertical direction.
component of centripetal acceleration = a * sin(angle)
Substituting the values, we get:
component of centripetal acceleration = 3.62g * sin(24.0°)
Calculating this:
component of centripetal acceleration = 3.62 * 9.8 m/s² * sin(24.0°)
component of centripetal acceleration ≈ 2.21g
Therefore, the magnitude of the component of the centripetal acceleration along the driver's upper body is approximately 2.21g.
(a) To find the speed of the car during the maneuver, we can use the centripetal acceleration formula:
a = v^2 / r
where a is the centripetal acceleration, v is the speed of the car, and r is the radius of the curve.
First, let's convert the radius from meters to kilometers:
229 m = 0.229 km
Now, we can rearrange the formula to solve for v:
v = sqrt(a * r)
Plugging in the values:
v = sqrt(3.62 * 0.229)
v ≈ 0.915 km/s
This answer is in km/s, but we need to convert it to km/h. Since there are 3600 seconds in an hour:
v = 0.915 * 3600
v ≈ 3294 km/h
Therefore, the speed of the car during the maneuver is approximately 3294 km/h.
(b) The component of the centripetal acceleration along the driver's upper body can be found using trigonometry. In this case, we need the component of the acceleration perpendicular to the banked curve. Since the driver's upper body is aligned with the normal direction to the curve, this perpendicular component is the same as the vertical component of the acceleration.
The vertical component of the acceleration can be found using the formula:
a_vertical = a * sin(θ)
where a is the centripetal acceleration and θ is the angle of the banked curve.
Plugging in the values:
a_vertical = 3.62 * sin(24.0°)
a_vertical ≈ 1.490 g
Therefore, the magnitude of the component of the centripetal acceleration along the driver's upper body is approximately 1.490 g.