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A test car starts from rest on a horizontal circular track of \(80-\mathrm{m}\) radius and increases its speed at a uniform rate to reach \(100 \mathrm{km} / \mathrm{h}\) in 10 seconds. Determine the magnitude \(a\) of the total acceleration of the car 8 seconds after the start.

Short Answer

Expert verified
The total acceleration after 8 seconds is approximately 6.78 m/s².

Step by step solution

01

Convert Units

First, convert the final speed from km/h to m/s. Given speed is 100 km/h. To convert to m/s, use the conversion factor: 1 km/h = 0.27778 m/s.\[ 100 \text{ km/h} = 100 \times 0.27778 = 27.78 \text{ m/s} \]
02

Calculate Tangential Acceleration

Since the car starts from rest and reaches 27.78 m/s in 10 seconds, we can calculate the tangential acceleration using the formula: \[ a_t = \frac{\Delta v}{\Delta t} = \frac{27.78 \text{ m/s} - 0 \text{ m/s}}{10 \text{ s}} = 2.778 \text{ m/s}^2 \]
03

Determine Velocity at 8 Seconds

To find the velocity at 8 seconds, use the formula: \[ v = u + a_t t \]where \( u \) is the initial velocity (0 m/s), \( a_t \) is the tangential acceleration, and \( t \) is 8 s.\[ v = 0 + 2.778 \times 8 = 22.22 \text{ m/s} \]
04

Calculate Centripetal Acceleration

Centripetal acceleration is given by \( a_c = \frac{v^2}{r} \). Using the velocity at 8 seconds, \( v = 22.22 \text{ m/s} \), and radius \( r = 80 \text{ m} \):\[ a_c = \frac{(22.22)^2}{80} = 6.17 \text{ m/s}^2 \]
05

Calculate Total Acceleration

The total acceleration is the vector sum of tangential and centripetal accelerations. Use the Pythagorean theorem:\[ a = \sqrt{a_t^2 + a_c^2} \]where \( a_t = 2.778 \text{ m/s}^2 \) and \( a_c = 6.17 \text{ m/s}^2 \).\[ a = \sqrt{(2.778)^2 + (6.17)^2} \approx 6.78 \text{ m/s}^2 \]

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Key Concepts

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

Circular Motion
Circular motion refers to the movement of an object along the circumference of a circle. This can occur when a vehicle, in this case, a test car, travels around a circular track. There are two crucial types of motions and forces to consider in circular motion: tangential (linear) and centripetal (radial). When dealing with circular motion, it's imperative to understand that the object changes its direction continuously, even if its speed is constant.
Generally, the path taken by the object is defined by the radius of the circle it follows. For our example, the test car moves on a track with an 80-meter radius. Having directional changes as the primary feature, circular motion uniquely combines both linear and radial components of motion.
Tangential Acceleration
Tangential acceleration is the rate of change of the linear speed of an object moving along a circular path. It occurs when there is a change in the speed of the object as it moves around the circle.
This type of acceleration is comparable to how an object increases its speed in a straight line. It is calculated using the formula for linear acceleration, which here is:
  • \(a_t = \frac{\Delta v}{\Delta t}\)
Here, \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time. For the test car, starting from rest and reaching 27.78 m/s in 10 seconds, the tangential acceleration is 2.778 m/s².
Tangential acceleration affects how rapidly the speed of the car increases as it travels along its circular path.
Centripetal Acceleration
Centripetal acceleration arises when an object is moving in a circular path and is directed towards the center of the circle. It is this acceleration that keeps the object in circular motion and is given by the formula:
  • \(a_c = \frac{v^2}{r}\)
Where \(v\) is the velocity of the object and \(r\) is the radius of the circle.
Using the velocity of the car at 8 seconds (22.22 m/s) and the radius of 80 meters, the centripetal acceleration is calculated to be 6.17 m/s².
Centripetal acceleration is not about changing the speed, but changing the direction of movement, maintaining the car's path along the circle. It is crucial for keeping any object in a confined circular trajectory.
Unit Conversion
Unit conversion is an essential skill in physics to ensure that all measurements are in the correct units for calculations. The most common conversions involve translating between the metric and imperial systems or converting between different metric units.
In the context of dynamics and this problem specifically, it's necessary to convert the car's speed from kilometers per hour to meters per second. This conversion helps standardize unit measurements in the calculations.
The conversion factor here is 1 km/h = 0.27778 m/s, and we use this to convert 100 km/h to 27.78 m/s. This step is vital because physics calculations need consistent units to yield accurate results. Proper unit conversion is a foundational part of mastering problem-solving in physics.

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