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When the bicycle passes point \(A\), it has a speed of \(6 \mathrm{~m} / \mathrm{s}\), which is increasing at the rate of \(\dot{v}=(0.5) \mathrm{m} / \mathrm{s}^{2}\). Determine the magnitude of its acceleration when it is at point \(A\).

Short Answer

Expert verified
The magnitude of acceleration of the bicycle at point \(A\) is \(0.5 \, m/s^2\).

Step by step solution

01

Understanding the given values

The given values are the speed \(v = 6 \, m/s\) of the bicycle at point \(A\) and the increasing rate of this speed \(\dot{v} = 0.5 \, m/s^2\). This rate of increase in speed is actually the tangential acceleration \(a_t\) of the bicycle.
02

Calculate the radial acceleration

The radial acceleration \(a_r\) or centripetal acceleration can be calculated using the formula: \(a_r = \frac{v^2}{r}\). However, since we are not given the radius of the bicycle's path, we cannot calculate the radial acceleration.
03

Determine the total acceleration

The total acceleration \(a\) is the vector sum of the radial acceleration \(a_r\) and the tangential acceleration \(a_t\). Its magnitude can be calculated according to the formula: \(a = \sqrt{{a_r}^2 + {a_t}^2}\). However, since we do not have a value for \(a_r\), the total acceleration in this case would be equal to the tangential acceleration \(a_t\), which is given as \(0.5 \, m/s^2\).

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Key Concepts

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

Understanding Tangential Acceleration
Tangential acceleration is the change in the speed of an object as it moves along a path. It measures how fast the object's speed is increasing or decreasing. If you've ever ridden a bicycle, think of it as pedaling harder to increase your speed.
The tangential acceleration is denoted by \( a_t \) and is related to the rate of change of speed \( \dot{v} \). For instance, in our original exercise, the bicycle's speed is increasing at a rate of \( 0.5 \, m/s^2 \). This is exactly what we refer to as its tangential acceleration.
  • Tangential acceleration changes the speed.
  • Measured in the same units as linear acceleration (\( m/s^2 \)).
  • Directly affects how quickly an object moves along a path.
Understanding this type of acceleration is crucial because it impacts how swiftly or slowly an object can traverse along its path.
Radial Acceleration: The Inward Pull
Radial acceleration, also known as centripetal acceleration, is crucial in understanding how objects move along curved paths. Imagine a bicycle moving along a circular path. While the speed changes due to tangential acceleration, radial acceleration keeps the bicycle from flying off the track.
The formula for radial acceleration is \( a_r = \frac{v^2}{r} \), where \( v \) is the speed and \( r \) is the radius of the path. While we didn't calculate it in our exercise, this inward acceleration ensures any object in circular motion stays on course.
  • Radial acceleration points toward the center of the circular path.
  • It's responsible for the "inward" pull that maintains the circular motion.
  • Measured in \( m/s^2 \), like other accelerations.
To sum up, radial acceleration is all about steering the object in a curved trajectory, ensuring it follows the intended path.
Total Acceleration: The Vector Sum
When discussing any object's acceleration in a curved path, it's essential to consider both tangential and radial accelerations. Total acceleration combines these two respective components into a single measure. This is called the vector sum.
In the case of our bicycle example, total acceleration would generally be calculated using \( a = \sqrt{{a_r}^2 + {a_t}^2} \). This formula combines both radial and tangential components to find the overall acceleration at any point on the path.
  • The vector sum accounts for both speed changes and direction changes.
  • Without radial acceleration information, it equals the tangential acceleration (if only one component is present).
  • Gives a comprehensive view of an object's motion on a path.
In our specific problem, since the radial component wasn't identifiable, the total acceleration simplified to just the tangential acceleration, \( 0.5 \, m/s^2 \). Always remember, real-world scenarios often require you to consider both components for an accurate depiction of motion.

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Most popular questions from this chapter

A ball \(A\) is thrown vertically upward from the top of a 30 -m-high building with an initial velocity of \(5 \mathrm{~m} / \mathrm{s}\). At the same instant another ball \(B\) is thrown upward from the ground with an initial velocity of \(20 \mathrm{~m} / \mathrm{s}\). Determine the height from the ground and the time at which they pass.

The rod \(O A\) rotates counterclockwise with a constant angular velocity of \(\dot{\theta}=5 \mathrm{rad} / \mathrm{s}\). 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 \(\theta=120^{\circ}\).

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The velocity of a particle is \(\mathbf{v}=\\{3 \mathbf{i}+(6-2 t) \mathbf{j}\\} \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds. If \(\mathbf{r}=\mathbf{0}\) when \(t=0\), determine the displacement of the particle during the time interval \(t=1 \mathrm{~s}\) to \(t=3 \mathrm{~s}\)

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