A racecar is initially travelling at 75 mph at point A as it enters the S-curve shown. In order to successfully traverse the curve, the racecar driver applies his brakes and decelerates uniformly between point A and B. Point B is located 750 ft down the track from point A. After point B, the racecar driver accelerates uniformly for 850 ft until he returns the vehicle’s speed to 75 mph at point C. If the total acceleration of the vehicle at point A and point C is 10 ft/s2, determine
the tangential acceleration at point A
the total acceleration and velocity at the car approaches point B
the radius of curvature of the track at point C.
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Kinematics of Particles - Curvilinear Motion
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To solve this problem, we need to break it down into different parts and analyze the motion of the racecar at each point. Let's go step by step:
1. Tangential acceleration at point A:
Tangential acceleration is the rate of change of speed. Since the racecar is decelerating uniformly, we can use the equation:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity vf is 0 mph (since the car stops), the initial velocity vi is 75 mph, and the acceleration a is what we need to find.
We are given that the total acceleration at point A and point C is 10 ft/s^2. Since tangential acceleration is a component of the total acceleration, we can write:
a = √(total acceleration^2 - radial acceleration^2)
Plugging in the values:
10 = √(a^2 - radial acceleration^2)
We need to find the tangential acceleration at point A, so we let radial acceleration = 0:
10 = √(a^2 - 0^2)
Simplifying, we get:
100 = a^2
Taking the square root, we find:
a = ± 10 ft/s^2
Since the racecar is decelerating, the tangential acceleration will be negative. Therefore, the tangential acceleration at point A is -10 ft/s^2.
2. Total acceleration and velocity at point B:
To find the total acceleration at point B, we can use the equation:
vf^2 = vi^2 + 2ad
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance.
In this case, we know the initial velocity vi is 75 mph, the final velocity vf is what we need to find, the acceleration a is -10 ft/s^2, and the distance d is 750 ft.
Converting 75 mph to ft/s, we get:
vi = 75 mph * (5280 ft/1 mile) * (1 hour/3600 seconds) = 110 ft/s
Plugging in the values into the equation:
vf^2 = (110 ft/s)^2 + 2(-10 ft/s^2)(750 ft)
vf^2 = 12100 ft^2/s^2 - 15,000 ft^2/s^2
vf^2 = -3900 ft^2/s^2
Since the velocity cannot be negative, we can deduce that the racecar is not able to reach point B.
Therefore, the total acceleration and velocity at the car approaches point B are undefined.
3. Radius of curvature of the track at point C:
To find the radius of curvature at point C, we can use the equation:
r = (v^2) / a
where r is the radius of curvature, v is the velocity, and a is the acceleration.
We are given that the velocity v at point C is 75 mph. Converting it to ft/s, we get:
v = 75 mph * (5280 ft/1 mile) * (1 hour/3600 seconds) = 110 ft/s
We are also given that the total acceleration at point C is 10 ft/s^2. Since this includes both the tangential and radial acceleration, we can use it directly in the equation.
Plugging in the values:
r = (110 ft/s)^2 / 10 ft/s^2
r = 12100 ft^2/s^2 / 10 ft/s^2
r = 1210 ft
Therefore, the radius of curvature of the track at point C is 1210 ft.