Anisha S. answered • 30d

Experienced Tutor for Middle/High School and College Students

From physics we know that circular acceleration = v^{2}/R, and that centripetal acceleration points inwards.

- First we need to find the speed of the car, v. Find the distance the car travels which is just the circumference of the track: 2π * 230m = 1445m. Speed = Distance/Time, so v = 1445m/52s = 27.79m/s. Now, to get the magnitude of the acceleration, we simply plug our known values into a = v
^{2}/R: 3.358m/s^{2}. - To find the x component of the velocity at point A, we have to take the cosine of angle A and multiply times the found velocity. cos(35)*27.79 = 22.765m/s. Velocity is tangential to the path of acceleration, so to determine the sign of the x-velocity, we draw a tangent line to point A (in the direction the car is moving) and see that it points left. Thus, the x component of the velocity at point A is -22.765m/s.
- Similar to 2, but now we take the sine. sin(35)*27.79 = 15.94m/s. The sign is positive because the tangential line at point A points in the upward direction.
- We do a similar process as 2 & 3, but now use the centripetal acceleration. cos(59) * 3.358m/s
^{2}= 1.729. To determine the sign for centripetal acceleration, remember that the acceleration vector points inward. Drawing the vector from Point B to the center of the track, we see that the vector points slightly left. Therefore, the x component of the acceleration vector is -1.729m/s^{2}. - Similar to 4. sin(59) * 3.358m/s
^{2}= 2.878. Drawing the vector from Point B to the center, we see that it points up. Therefore, the y component of the acceleration vector is positive 2.878m/s^{2}. - Increasing. If you draw a line tangent to point B and some other lines tangent to the areas around B, you'll see that the slopes of the lines are increasing in the y-direction.