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.

To determine the tangential acceleration at point A, we need to find the change in velocity from point A to point B. We can calculate this using the equation:

Δv = vf - vi

where:
Δv is the change in velocity
vf is the final velocity
vi is the initial velocity

In this case, the racecar's initial velocity is 75 mph, and it decelerates uniformly until reaching point B. Therefore, the final velocity at point B is 0 mph.

Converting the velocities to the same unit (feet per second) by multiplying by the conversion factor of 1.467, we have:
vi = 75 mph * 1.467 = 110.025 ft/s
vf = 0 mph * 1.467 = 0 ft/s

Substituting these values into the equation, we get:

Δv = 0 ft/s - 110.025 ft/s
Δv = -110.025 ft/s

Since the racecar is decelerating, the change in velocity is negative. Therefore, the tangential acceleration at point A is equal to the change in velocity divided by the distance between point A and point B.

Let's calculate the tangential acceleration at point A:

Distance between point A and point B = 750 ft

tangential acceleration at point A = Δv / distance
tangential acceleration at point A = (-110.025 ft/s) / (750 ft)
tangential acceleration at point A ≈ -0.147 ft/s²

Moving on to the next part:

To determine the total acceleration and velocity as the car approaches point B, we need to consider that the racecar is decelerating from point A to point B and then accelerating from point B to point C. The total acceleration at point B is the sum of the tangential acceleration at point A and the acceleration due to the change in velocity.

Given:
Total acceleration at point A = -10 ft/s²
Distance between point B and point C = 850 ft

We can calculate the acceleration due to the change in velocity using the kinematic equation:

vf² = vi² + 2ad

where:
vf is the final velocity
vi is the initial velocity
a is the acceleration
d is the distance

Since the final velocity at point C is 75 mph, let's convert it to feet per second:

vf = 75 mph * 1.467 = 110.025 ft/s

The initial velocity at point B is 0 ft/s.

Substituting these values into the kinematic equation, we get:

vf² = vi² + 2ad
(110.025 ft/s)² = (0 ft/s)² + 2a(850 ft)
12105.625 ft²/s² = 1700a ft

Simplifying the equation:

12105.625 = 1700a

Solving for a, we divide both sides by 1700:

a = 12105.625 / 1700
a ≈ 7.119 ft/s²

Now, we can find the total acceleration at point B by summing the tangential acceleration at point A and the acceleration due to the change in velocity:

total acceleration at point B = tangential acceleration at point A + acceleration due to change in velocity
total acceleration at point B ≈ -0.147 ft/s² + 7.119 ft/s²
total acceleration at point B ≈ 6.972 ft/s²

As for the velocity at point B, it remains to be determined. To find it, we can use the equation:

vf = vi + at

where:
vf is the final velocity
vi is the initial velocity
a is the acceleration
t is the time

Since we don't have the time information, we can't solve for the exact velocity at point B. However, we can still determine the trend of the velocity at point B. Since the car is decelerating uniformly from point A to point B, the velocity at point B will be between the initial velocity at point A (110.025 ft/s) and 0 ft/s.

For the final part:

To determine the radius of curvature of the track at point C, we need to consider that when a car is moving along a curved path, the net force acting on it provides the necessary centripetal force to make the car follow the curve. The centripetal force is given by the formula:

Fc = m * ac

where:
Fc is the centripetal force
m is the mass of the car
ac is the centripetal acceleration

Since we don't have the mass of the car, we can calculate the centripetal acceleration using the tangential acceleration at point C, which we can find by using the equation:

ac = v² / r

where:
ac is the centripetal acceleration
v is the velocity at point C
r is the radius of curvature

We know that the velocity at point C is 75 mph, which we can convert to feet per second:

v = 75 mph * 1.467 = 110.025 ft/s

Now, we need to solve for r. Rearranging the equation, we get:

r = v² / ac

Substituting the values we have:

r = (110.025 ft/s)² / (10 ft/s²)
r = 12105.625 ft²/s² / 10 ft/s²
r = 1210.5625 ft

Therefore, the radius of curvature of the track at point C is approximately 1210.56 ft.