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Determine the maximum constant speed a race car can have if the acceleration of the car cannot exceed \(7.5 \mathrm{m} / \mathrm{s}^{2}\) while rounding a track having a radius of curvature of \(200 \mathrm{m}\).

Short Answer

Expert verified
The maximum constant speed a race car can have is approximately \(122.47 \, m/s\).

Step by step solution

01

Write down the Centripetal Acceleration Formula

The centripetal acceleration formula is \(a = \frac{v^2}{r}\), where \(a\) is the centripetal acceleration, \(v\) is the speed, and \(r\) is the radius.
02

Rearrange the Formula to Solve for Speed

Rearrange the formula to solve for speed. Multiplying both sides of the equation by \(r\) gives \(ar = v^2\). Then, take the square root of both sides to solve for \(v\), resulting in \(v = \sqrt{ar}\).
03

Substitute the Given Values into the Formula

Substitute the given maximum acceleration \(a = 7.5 \, m/s^2\) and radius of curvature \(r = 200 \, m\) into the formula: \(v = \sqrt{(7.5 \, m/s^2)(200 \, m)}\).
04

Calculate the Maximum Speed

Calculate the maximum speed. Following the multiplication and square root operation, \(v_{max}\approx 122.47 \, m/s\) provided that the car maintains the maximum allowed acceleration while rounding the track.

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Key Concepts

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

Race Car Dynamics
When discussing race car dynamics, one crucial element is understanding how a car moves on a track. The dynamics of a race car involves examining its motion, including velocity and acceleration, especially when turning around curves.

Race cars need to maintain high speeds while ensuring stability and control. Various forces act on the vehicle as it moves, such as the friction between the tires and the track, downforce generated by aerodynamic components, and the centripetal force necessary for navigating turns.
  • Friction helps in stabilizing the car by preventing slippage.
  • Aerodynamics play a role in keeping the car glued to the track.
  • Centripetal force aids in making the car turn smoothly without flying off the curve.
By understanding these forces, car designers can make informed decisions to enhance the car's performance and safety.
Radius of Curvature
The radius of curvature is a key factor when calculating how a car performs on a curved track. It represents the radius of an imaginary circle that matches the curve of the track at that specific point. The radius of curvature affects how tight a corner is and influences the maximum speed a vehicle can maintain while rounding the curve.

A smaller radius indicates a tighter turn, which typically requires slower speeds to maintain control. Conversely, a larger radius curve can be taken at higher speeds safely. The relation between the radius of curvature and centripetal acceleration is crucial for high-speed maneuvering. It allows the car to adjust its velocity precisely to maintain the trajectory along the curve.
  • Larger radius = gentler turn, potential for higher speeds.
  • Smaller radius = tighter turn, necessitates lower speeds.
Maximum Speed Calculation
Calculating the maximum speed a race car can achieve while safely turning a curve involves using the centripetal acceleration formula. This formula is essential because it connects acceleration, speed, and the radius of curvature in a tangible way.

Here’s a step-by-step breakdown:
  • Identify the maximum allowed acceleration (in this case, 7.5 m/s²).
  • Use the radius of curvature provided (200 m for this exercise).
  • Apply the centripetal acceleration formula: \(a = \frac{v^2}{r}\), where \(v\) is the speed you're solving for.
  • To find \(v\), rearrange it to \(v = \sqrt{ar}\), then substitute the given values.
  • Solve to find \(v_{max} \), which calculates to approximately 122.47 m/s.
Understanding this calculation helps ensure the vehicle stays on the track without skidding or losing control.
Physics of Motion
The physics of motion is fundamental in understanding how race cars, or any objects for that matter, move and react in different environments. Centripetal acceleration is a concept in physics that helps us describe this motion, especially when an object is following a curved path.

When a race car rounds a curve, it continuously changes direction, making acceleration occur even if the speed remains constant. This change of direction is where centripetal force comes into play. Centripetal acceleration helps to keep the race car on its curved path.
  • Newton’s second law of motion connects force, mass, and acceleration: \(F = ma\). This principle helps in understanding that the centripetal force required to keep a car in a turn is \(F = \frac{mv^2}{r}\).
  • Without sufficient centripetal force and acceleration, the car would move off tangent to the curve.
By appreciating these physics concepts, drivers and engineers can optimize the performance and safety of race cars on any track.

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Most popular questions from this chapter

The rod \(O A\) rotates counterclockwise with a constant angular velocity of \(\dot{\theta}=5\) rad / s. Two pin-connected slider blocks, located at \(B,\) move freely on \(O A\) and the curved rod whose shape is a limaçon described by the equation \(r=100(2-\cos \theta) \mathrm{mm} .\) Determine the speed of the slider blocks at the instant \(\theta=120^{\circ}\).

A particle is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by \(\theta=\cos 2 t,\) where \(\theta\) is in radians and \(t\) is in seconds. Determine the magnitude of the acceleration of the particle when \(\theta=30^{\circ}\).

The race car starts from rest and travels along a straight road until it reaches a speed of \(26 \mathrm{m} / \mathrm{s}\) in \(8 \mathrm{s}\) as shown on the \(v-t\) graph. The flat part of the graph is caused by shifting gears. Draw the \(a-t\) graph and determine the maximum acceleration of the car.

The acceleration of a particle traveling along a straight line is \(a=\frac{1}{4} s^{1 / 2} \mathrm{m} / \mathrm{s}^{2},\) where \(s\) is in meters. If \(v=0\) \(s=1 \mathrm{m}\) when \(t=0,\) determine the particle's velocity at \(s=2 \mathrm{m}\).

The velocity of a particle traveling along a straight line is \(v=\left(3 t^{2}-6 t\right) \mathrm{ft} / \mathrm{s},\) where \(t\) is in seconds. If \(s=4 \mathrm{ft}\) when \(t=0,\) determine the position of the particle when \(t=4\) s. What is the total distance traveled during the time interval \(t=0\) to \(t=4 \mathrm{s} ?\) Also, what is the acceleration when \(t=2 \mathrm{s} ?\)

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