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Starting from rest at time \(t=0\), a circus stunt man drives a motorbike on a horizontal circular track of radius \(10.0 \mathrm{~m}\). His speed is given by \(v=c t^{2}\), where \(c=1.00 \mathrm{~m} / \mathrm{s}^{3}\). At \(t=2.00 \mathrm{~s}\), what is the angle between his (total) acceleration vector and his radial acceleration vector?

Short Answer

Expert verified
The angle is approximately \( 68.2^\circ \).

Step by step solution

01

Identify the velocity equation

The velocity of the stunt man is given by the equation \( v = c t^2 \), where \( c = 1.00 \, \text{m/s}^3 \). At \( t = 2.00 \, \text{s} \), substitute to find the velocity: \( v = 1.00 \, (2.00)^2 = 4.00 \, \text{m/s} \).
02

Calculate the radial acceleration

The radial acceleration is given by \( a_r = \frac{v^2}{r} \), where \( r = 10.0 \, \text{m} \). Substitute the values \( v = 4.00 \, \text{m/s} \) and \( r = 10.0 \, \text{m} \) to calculate \( a_r \): \ \( a_r = \frac{(4.00)^2}{10.0} = 1.60 \, \text{m/s}^2 \).
03

Calculate the tangential acceleration

The tangential acceleration is the rate of change of the speed \( v \) with respect to time \( t \). \( a_t = \frac{dv}{dt} = \frac{d}{dt}(c t^2) \). Compute the derivative: \ \( a_t = 2c t = 2(1.00) \times 2.00 = 4.00 \, \text{m/s}^2 \).
04

Calculate the total acceleration

The total acceleration \( a_{total} \) is the vector sum of radial and tangential accelerations. Use Pythagorean theorem: \ \( a_{total} = \sqrt{a_r^2 + a_t^2} = \sqrt{(1.60)^2 + (4.00)^2} = \sqrt{2.56 + 16.00} = \sqrt{18.56} = 4.31 \, \text{m/s}^2 \).
05

Calculate the angle between the accelerations

The angle \( \theta \) between \( a_{total} \) and \( a_r \) can be found using: \ \( \tan \theta = \frac{a_t}{a_r} = \frac{4.00}{1.60} = 2.5 \). Use inverse tangent function to find \( \theta \): \ \( \theta = \tan^{-1}(2.5) \approx 68.2^\circ \).

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Key Concepts

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

Radial Acceleration
When an object moves in a circular path, even if it maintains constant speed, its direction is constantly changing. This change in direction is due to radial or centripetal acceleration. In simple terms, radial acceleration acts towards the center of the circular path.
This kind of acceleration is responsible for keeping the object on its curved path. To find radial acceleration, we use the formula:
  • \( a_r = \frac{v^2}{r} \)
Here, \( v \) represents the object's velocity, and \( r \) is the radius of the circle. For example, if the velocity is \( 4.00 \, \text{m/s} \) and the circle's radius is \( 10.0 \, \text{m} \), the radial acceleration will be \( 1.60 \, \text{m/s}^2 \). Remember, radial acceleration is always directed inward, towards the circle’s center.
Tangential Acceleration
While radial acceleration focuses on changing the direction of movement, tangential acceleration looks at changing the speed along the tangent of the path. It's a linear acceleration and occurs when the speed of the object increases or decreases.
To find tangential acceleration, we use the derivative of the velocity with respect to time:
  • \( a_t = \frac{dv}{dt} \)
For the stuntman, whose velocity is given by \( v = c t^2 \), taking the derivative results in:
  • \( a_t = 2c t \)
Substituting \( c = 1.00 \, \text{m/s}^3 \) and \( t = 2.00 \, \text{s} \), the tangential acceleration becomes \( 4.00 \, \text{m/s}^2 \).Unlike radial acceleration, tangential acceleration can point forward or backward, depending on whether the object's speed is increasing or decreasing along the path.
Total Acceleration
Total acceleration is the combination of both radial and tangential accelerations. It is a vector quantity, meaning it has both magnitude and direction.
To calculate the magnitude of total acceleration, use the Pythagorean theorem as both components are perpendicular to each other:
  • \( a_{\text{total}} = \sqrt{a_r^2 + a_t^2} \)
Given \( a_r = 1.60 \, \text{m/s}^2 \) and \( a_t = 4.00 \, \text{m/s}^2 \), this results in:
  • \( a_{\text{total}} = 4.31 \, \text{m/s}^2 \)
To find the angle between the total acceleration and radial acceleration, we use:
  • \( \tan \theta = \frac{a_t}{a_r} \)
This equation gives the angle \( \theta = 68.2^\circ \) between the acceleration vectors. This angle tells us how the total acceleration vector is oriented relative to the radial direction, providing a clearer picture of motion dynamics.

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