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Chapter 11: Problem 143

A race car enters the circular portion of a track that has a radius of \(70 \mathrm{m}\). When the car enters the curve at point \(P,\) it is travelling with a speed of \(120 \mathrm{km} / \mathrm{h}\) that is increasing at \(5 \mathrm{m} / \mathrm{s}^{2} .\) Three seconds later, determine the \(x\) and \(y\) components of velocity and acceleration of the car.

Short Answer

Expert verified
Velocity components: \(v_x = 48.33\, \mathrm{m/s}, v_y = 0\, \mathrm{m/s}\). Acceleration components: \(a_x = 5\, \mathrm{m/s}^2, a_y = 33.38\, \mathrm{m/s}^2\) towards center.

Step by step solution

01

Convert the Initial Speed

The initial speed of the car is given in km/h. First, convert this speed to meters per second (m/s). Given:Initial speed = 120 km/hConversion: 1 km/h = 1000 m/3600 s, hence 120 km/h = \(\frac{120 \times 1000}{3600}\, \mathrm{m/s} = 33.33\, \mathrm{m/s}\).
02

Calculate the Tangential Velocity After 3 Seconds

Since the car's speed is increasing at a rate of 5 m/s虏, we need to find the tangential velocity after 3 seconds. Tangential acceleration \(a_t = 5\, \mathrm{m/s}^2\).Using the formula \(v_f = v_i + a_t \times t\), where \(v_i\) is the initial velocity, we find:\[v_f = 33.33 \, \mathrm{m/s} + 5 \, \mathrm{m/s}^2 \times 3\, \mathrm{s} = 48.33 \, \mathrm{m/s}\].
03

Calculate the Radial Acceleration

The radial (centripetal) acceleration can be calculated using the final tangential speed.Centripetal acceleration \(a_r = \frac{v^2}{r}\), where \(v = 48.33\, \mathrm{m/s}\) and radius \(r = 70\, \mathrm{m}\).\[a_r = \frac{(48.33 \, \mathrm{m/s})^2}{70\, \mathrm{m}}\approx 33.38 \, \mathrm{m/s}^2\].
04

Determine the Velocity Components

The car is moving tangentially in a circular path. After 3 seconds, the velocity components are:- In the direction of motion (tangential), the velocity is \(v = 48.33 \, \mathrm{m/s}\).- Perpendicular to the radius (centripetal), it has no component, so the x-component \(v_x = v\cos(\theta)\) and y-component \(v_y = v\sin(\theta)\). Without specific angles, assume motion along x-axis: Thus, \(v_x = 48.33 \, \mathrm{m/s}\) and \(v_y = 0 \, \mathrm{m/s}\).
05

Determine the Acceleration Components

The acceleration components include both tangential and radial components:- Tangential acceleration (\(a_{tx}\)): 5 m/s虏.- Radial (centripetal) acceleration (\(a_{ry}\)): equivalent to \(a_r = 33.38 \, \mathrm{m/s}^2\).Consider net acceleration:- \(a_x = a_{tx} = 5 \, \mathrm{m/s}^2\)- \(a_y = a_{ry} = 33.38 \, \mathrm{m/s}^2\) (acting toward the center of the track).Hence, \(a_x = 5\, \mathrm{m/s}^2\) and \(a_y = 33.38\, \mathrm{m/s}^2\) towards the center of curvature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Acceleration
In circular motion, centripetal acceleration is crucial to keep an object moving along a curved path without veering off. This type of acceleration acts inward, towards the center of the circular path. It's important to understand that centripetal acceleration does not increase the speed of the object. Instead, it constantly changes the direction of the velocity, which is essential for maintaining circular motion.
  • Formula: The centripetal acceleration (\(a_r\)) can be determined using the formula \(a_r = \frac{v^2}{r}\), where \(v\) is the speed of the moving object and \(r\) is the radius of the circular path.
  • Application: In our exercise, the car鈥檚 speed after 3 seconds is \(48.33 \, \mathrm{m/s}\), and with a radius of \(70 \, \mathrm{m}\), the centripetal acceleration is calculated to be approximately \(33.38 \, \mathrm{m/s}^2\).
  • Direction: Always remember that centripetal acceleration is directed toward the center of the arc or circle that the object is traversing.
Understanding centripetal acceleration helps explain why objects in circular motion need to keep turning, preventing them from flying outward, which is often misunderstood. This concept is critical when analyzing movement on curved tracks like our race car example.
Tangential Acceleration
Tangential acceleration occurs when the speed of an object along its path increases or decreases. It acts in the same direction as the motion along the circle's tangent line.
  • Concept: Unlike centripetal acceleration, tangential acceleration (\(a_t\)) directly changes the speed of the object as it moves along the path.
  • Formula: It is given by the relationship \(v_f = v_i + a_t \times t\), where \(v_f\) is the final speed, \(v_i\) is the initial speed, \(a_t\) is the tangential acceleration, and \(t\) is the time.
  • Example: In our scenario, the car's tangential acceleration was \(5 \, \mathrm{m/s}^2\), leading to an increase in speed from \(33.33 \, \mathrm{m/s}\) to \(48.33 \, \mathrm{m/s}\) over 3 seconds.
Knowing how tangential acceleration works is essential for understanding how velocity changes in uniform circular motion. It helps explain how vehicles can speed up or slow down while navigating a curve, which is reflected in our race car problem.
Velocity Components
Velocity components in circular motion are broken down into two directions: one along the path (tangential) and one perpendicular to the path (radial). Properly understanding these components is vital for solving problems involving motion on a curved path.
  • Tangential component (\(v_x\)): This is the velocity along the direction of motion. In the given exercise, since no specific angle is provided, it is assumed that the car travels predominantly along the x-axis with a speed of \(48.33 \, \mathrm{m/s}\).
  • Radial component (\(v_y\)): Because the car moves in a circle and follows the x-axis assumption, the radial velocity component is zero (\(v_y = 0 \, \mathrm{m/s}\)).
  • Combining components: These components help in analyzing and predicting the movement of objects along curved paths, such as finding the resultant velocity or the effect of external forces.
The velocity components give a more detailed picture of how an object moves and are foundational for calculations in dynamics and kinematics. In the context of the problem, they help calculate the car's velocity as it moves around the circular track.

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Most popular questions from this chapter

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