During a high-speed chase, a 2400-lb sports car traveling
at a speed of 100 mi/h just loses contact with the road as it
reaches the crest A of a hill. a) Determine the radius of
curvature r of the vertical profile of the road at A. b)
Using the value of r found in a part a), determine the
force exerted on a 160-lb driver by the seat of his 3100-lb
car as the car, traveling at a constant speed of 50 mi/h,
passes through A.
a) To determine the radius of curvature, we need to calculate the centripetal acceleration experienced by the car as it loses contact with the road.
The centripetal acceleration, a_c, is given by the formula:
a_c = v^2 / r
where v is the speed of the car and r is the radius of curvature.
Given:
Mass of the car, m = 2400 lb
Speed of the car, v = 100 mi/h
First, we need to convert the mass of the car from pounds to slugs. Since 1 slug is equal to the mass that is accelerated at 1 ft/s^2 when a force of 1 lb is applied, we can convert pounds to slugs by dividing by the acceleration due to gravity, g, which is 32.17 ft/s^2.
m = 2400 lb / (32.17 ft/s^2)
m = 74.6 slugs
Next, we need to convert the speed of the car from miles per hour to feet per second. Since 1 mile is equal to 5280 feet and 1 hour is equal to 3600 seconds, we can convert miles per hour to feet per second by multiplying by 5280/3600.
v = 100 mi/h * (5280 ft / 3600 s)
v = 146.67 ft/s
Now we can calculate the centripetal acceleration:
a_c = (146.67 ft/s)^2 / r
a_c = 21486.67 ft^2/s^2 / r
Since the car loses contact with the road at the crest of the hill, the centripetal acceleration is equal to the gravitational acceleration, g.
a_c = g
Therefore, we can set the equation equal to solve for the radius of curvature, r:
21486.67 ft^2/s^2 / r = 32.17 ft/s^2
Solving for r:
r = 21486.67 ft^2/s^2 / 32.17 ft/s^2
r = 668.18 ft
Therefore, the radius of curvature of the vertical profile of the road at point A is approximately 668.18 ft.
b) To determine the force exerted on the driver by the seat of the car, we need to calculate the net force acting on the car as it passes through point A.
The net force, F_net, is given by the equation:
F_net = m * a_c
where m is the total mass of the car and a_c is the centripetal acceleration.
Given:
Mass of the driver, m_driver = 160 lb
Mass of the car, m_car = 3100 lb
Speed of the car, v = 50 mi/h
Converting the mass of the driver and the car from pounds to slugs:
m_driver = 160 lb / 32.17 ft/s^2 = 4.98 slugs
m_car = 3100 lb / 32.17 ft/s^2 = 96.27 slugs
Converting the speed of the car from miles per hour to feet per second:
v = 50 mi/h * (5280 ft / 3600 s) = 73.33 ft/s
Using the previously calculated radius of curvature, r = 668.18 ft, we can now calculate the centripetal acceleration:
a_c = v^2 / r
a_c = (73.33 ft/s)^2 / 668.18 ft
a_c = 8.05 ft/s^2
Now we can calculate the net force acting on the car:
F_net = (m_car + m_driver) * a_c
F_net = (96.27 slugs + 4.98 slugs) * 8.05 ft/s^2
F_net = 100.25 slugs * 8.05 ft/s^2
F_net = 808.26 lb
Therefore, the force exerted on the driver by the seat of the car as it passes through point A is approximately 808.26 lb.