Final answer:
The magnitude of the car's acceleration is 2.36 m/s^2. The x component of the car's velocity at point A is 20.33 m/s, while the y component is 9.18 m/s. The x component of the car's acceleration at point B is 1.01 m/s^2, while the y component is 1.97 m/s^2.
Step-by-step explanation:
1) The magnitude of the car's acceleration can be calculated using the equation a = v^2/r, where v is the velocity and r is the radius of the circular track. In this case, the velocity can be determined by dividing the circumference of the track by the time it takes to complete one lap. The circumference of the track is given by 2πr, so the velocity is (2πr)/t. Plugging in the given values, we have v = (2π * 220 m) / (61 s) = 22.85 m/s. Now we can calculate the magnitude of the acceleration: a = (22.85 m/s)^2 / 220 m = 2.36 m/s^2.
2) The x component of the car's velocity can be determined by multiplying the magnitude of the velocity by the cosine of the angle θA. We have vx = v × cos(θA) = 22.85 m/s × cos(23°) = 20.33 m/s.
3) The y component of the car's velocity can be determined by multiplying the magnitude of the velocity by the sine of the angle θA. We have vy = v × sin(θA) = 22.85 m/s × sin(23°) = 9.18 m/s.
4) The x component of the car's acceleration at point B can be determined using the same formula as in question 2, but with the angle θB. We have ax = a × cos(θB) = 2.36 m/s^2 v cos(57°) = 1.01 m/s^2.
5) The y component of the car's acceleration at point B can be determined using the same formula as in question 3, but with the angle θB. We have ay = a ×sin(θB) = 2.36 m/s^2 v sin(57°) = 1.97 m/s^2.