Math Problem Statement

A car drives past a police radar.

Calculate the amount of acceleration of the car if the police radar measures = 5.1 m/s², an angular velocity = -0.22 rad/s when the car is at distance r = 56.34 m and the angle is θ = 42°.

Solution

We are given the following information:

  • Acceleration measured by radar: a=5.1m/s2a = 5.1 \, \text{m/s}^2
  • Angular velocity: ω=0.22rad/s\omega = -0.22 \, \text{rad/s}
  • Distance (radius): r=56.34mr = 56.34 \, \text{m}
  • Angle: θ=42\theta = 42^\circ

To calculate the acceleration of the car, we need to consider both the tangential and radial components of acceleration in polar coordinates.

Step 1: Components of acceleration

In polar coordinates, acceleration has two components:

  1. Tangential acceleration ata_t: This is the component of acceleration in the direction of motion. It is given by: at=rαa_t = r \cdot \alpha where α\alpha is the angular acceleration.

  2. Radial acceleration ara_r: This is the component of acceleration directed towards or away from the center of rotation. It is given by: ar=rω2a_r = r \cdot \omega^2 where ω\omega is the angular velocity.

Step 2: Calculate the radial acceleration

Given that r=56.34mr = 56.34 \, \text{m} and ω=0.22rad/s\omega = -0.22 \, \text{rad/s}, we can calculate the radial acceleration: ar=rω2=56.34m(0.22rad/s)2a_r = r \cdot \omega^2 = 56.34 \, \text{m} \cdot (-0.22 \, \text{rad/s})^2 ar=56.34(0.0484)2.73m/s2a_r = 56.34 \cdot (0.0484) \approx 2.73 \, \text{m/s}^2

Step 3: Calculate the tangential acceleration

Now, we know that the total acceleration of the car aa is a combination of both radial and tangential acceleration components. The total acceleration is measured as 5.1m/s25.1 \, \text{m/s}^2, so we can use the following relation: a2=ar2+at2a^2 = a_r^2 + a_t^2 Substitute the known values: (5.1)2=(2.73)2+at2(5.1)^2 = (2.73)^2 + a_t^2 26.01=7.45+at226.01 = 7.45 + a_t^2 at2=26.017.45=18.56a_t^2 = 26.01 - 7.45 = 18.56 at=18.564.31m/s2a_t = \sqrt{18.56} \approx 4.31 \, \text{m/s}^2

Step 4: Conclusion

The tangential acceleration of the car is approximately 4.31m/s24.31 \, \text{m/s}^2, and the radial acceleration is approximately 2.73m/s22.73 \, \text{m/s}^2.

The total acceleration is the vector sum of these components, which matches the given 5.1m/s25.1 \, \text{m/s}^2.

Would you like more details or have any questions about the calculation?

Related Questions:

  1. How does the angular velocity affect the radial acceleration?
  2. What is the role of the angle θ\theta in this calculation?
  3. How can the tangential and radial accelerations be used to calculate the total acceleration?
  4. What would happen if the angular velocity increased in this scenario?
  5. How can you derive angular acceleration from the given information?

Tip: Understanding the difference between radial and tangential acceleration is crucial when analyzing motion in polar coordinates.

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Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Acceleration
Angular Velocity
Kinematics

Formulas

a_r = r * ω^2
a^2 = a_r^2 + a_t^2

Theorems

Radial and Tangential Components of Acceleration

Suitable Grade Level

Grades 10-12