The angle θA = 35° above the x axis, and the angle θB = 59° below the x axis. 1)What is the magnitude of the car’s acceleration?

Question

A car is traveling around a horizontal circular track with radius r = 230 m as shown. It takes the car t = 52 s to go around the track once. The angle θ_A = 35° above the x axis, and the angle θ_B = 59° below the x axis.

1)What is the magnitude of the car’s acceleration?

2)What is the x component of the car’s velocity when it is at point A

3)What is the y component of the car’s velocity when it is at point A

4)What is the x component of the car’s acceleration when it is at point B

5)What is the y component of the car’s acceleration when it is at point B

6)

As the car passes point B, the y component of its velocity is(choose one)

increasing

constant

decreasing

Answer 1

From physics we know that circular acceleration = v²/R, and that centripetal acceleration points inwards.

1. 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 = v²/R: 3.358m/s².
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.
3. 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.
4. We do a similar process as 2 & 3, but now use the centripetal acceleration. cos(59) * 3.358m/s²= 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/s².
5. Similar to 4. sin(59) * 3.358m/s²= 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/s².
6. 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.