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Chapter 12: Problem 135

Starting from rest, a bicyclist travels around a horizontal circular path, \(\rho=10 \mathrm{~m},\) at a speed of \(v=\left(0.09 t^{2}+0.1 t\right) \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds. Determine the magnitudes of his velocity and acceleration when he has traveled \(s=3 \mathrm{~m}\),

Short Answer

Expert verified
The magnitude of the velocity and acceleration of the bicyclist when he has traveled a total distance of 3m will be found by following the procedure outlined above. The first step is to find the time when the bicyclist has traveled 3m, then use that time to find the velocity. Subsequently, differentiate the velocity function to get the linear acceleration, and calculate the centripetal acceleration using the speed and radius. The total acceleration is then the vector sum of these two acceleration components.

Step by step solution

01

Find the time when the bicyclist has traveled a distance of 3m

From the equation of motion, distance traveled \(s\) is equivalent to the integral of the velocity equation over time. Represented as \(s = \int v dt\). Therefore, we integrate the velocity function \(v\) over time \(t\) to get \(s = \int (0.09t^2 + 0.1t) dt\). Solving the above equation when \(s = 3m\) will give us the value of time \(t\).
02

Find the magnitude of velocity at this time

After getting the value of \(t\) from step 1, we substitute it into our velocity function \(v = (0.09t^2 + 0.1t)\) to find the value of velocity at \(t\) seconds.
03

Find the linear acceleration at this time

The linear or tangential acceleration is simply the derivative of the velocity function. Hence we differentiate the velocity \(v\) to get the acceleration \(a = dv/dt\). Then substitute \(t\) from Step 1 into the acceleration equation to get the linear acceleration at that time.
04

Find the centripetal acceleration at this time

The centripetal acceleration in circular motion is given by \(a_c = v^2/蟻\). Using the velocity from Step 2 and the radius \(蟻 = 10m\), we can find centripetal acceleration.
05

Compute the total acceleration at this time

The total acceleration of the bicyclist in circular motion is the vector sum of the linear acceleration and the centripetal acceleration. This can be given as \(a = \sqrt{{a_t}^2 + {a_c}^2}\), where \(a_t\) is the linear acceleration and \(a_c\) is the centripetal acceleration.

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Key Concepts

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

Velocity Calculation
Velocity is a crucial concept when studying circular motion, as it describes how fast an object is traveling along a path. In this exercise, the velocity of the bicyclist is provided as a function of time:
  • \( v = 0.09t^2 + 0.1t \).
To find the bicyclist's velocity at a specific time, we plug the time value into this equation. This helps us understand how the speed changes as time progresses.
Velocity in circular motion not only has magnitude but also direction, constantly changing as the object moves along the path. However, this exercise focuses on the magnitude. Determining the time when the bicyclist has traveled a certain distance requires integrating the velocity function, giving us a better grasp of his journey along the circular path.
Linear Acceleration
The concept of linear acceleration explains how quickly the velocity of an object changes over time. In this scenario, we find the linear acceleration by differentiating the velocity function with respect to time:
  • \( a = \frac{dv}{dt} \).
The process of differentiation allows us to see how the slope of the velocity-time graph changes. This gives us the rate at which the bicyclist speeds up or slows down.
Once we obtain the expression for acceleration, substituting the time found earlier will yield the linear acceleration at that particular moment. It's essential to understand this concept because linear acceleration is what gets the bicyclist moving faster along his path.
Centripetal Acceleration
Centripetal acceleration is vital in circular motion, directing an object inward along its circular path. It keeps the object moving in a circular loop. For this exercise, we use:
  • \( a_c = \frac{v^2}{\rho} \),
where \( \rho \) represents the radius of the circular path. Centripetal acceleration depends on both the velocity at that moment and the radius.
To find it, we take the velocity computed and substitute it into the centripetal formula, acknowledging the role velocity plays in keeping the bicyclist on track. This inward acceleration does not change the speed but constantly changes the direction toward the center.
Integration in Physics
Integration is a powerful tool in physics, essential for solving problems where functions depend on continuous change. Here, it helps find the distance traveled over a period by integrating the velocity function.
  • \( s = \int v \, dt \).
This operation accumulates the area under the velocity-time curve to give us the total distance.
Understanding integration in the context of circular motion helps visualize how the cyclist's travel depends on continuous velocity change. It bridges the instantaneous velocity at each moment with the total path covered, offering insights into the dynamics of circular motion. Integration simplifies complex relationships into manageable forms, allowing clear insights into how movements evolve over time.

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Most popular questions from this chapter

A boy throws a ball straight up from the top of a \(12-\mathrm{m}\) high tower. If the ball falls past him \(0.75 \mathrm{~s}\) later, determine the velocity at which it was thrown, the velocity of the ball when it strikes the ground, and the time of flight.

The pin follows the path described by the equation \(r=(0.2+0.15 \cos \theta) \mathrm{m} .\) At the instant \(\theta=30^{\circ},\) \(\dot{\theta}=0.7 \mathrm{rad} / \mathrm{s}\) and \(\ddot{\theta}=0.5 \mathrm{rad} / \mathrm{s}^{2} .\) Determine the magnitudes of the pin's velocity and acceleration at this instant. Neglect the size of the pin.

Two rockets start from rest at the same elevation. Rocket \(A\) accelerates vertically at \(20 \mathrm{~m} / \mathrm{s}^{2}\) for \(12 \mathrm{~s}\) and then maintains a constant speed. Rocket \(B\) accelerates at \(15 \mathrm{~m} / \mathrm{s}^{2}\) until reaching a constant speed of \(150 \mathrm{~m} / \mathrm{s}\). Construct the \(a-t, v-t,\) and \(s-t\) graphs for each rocket until \(t=20 \mathrm{~s}\). What is the distance between the rockets when \(t=20 \mathrm{~s}\) ?

A particle is traveling with a velocity of \(\mathbf{v}=\left\\{3 \sqrt{t} e^{-0.2 t} \mathbf{i}+4 e^{-0.8 t} \mathbf{j}\right\\} \mathrm{m} / \mathrm{s},\) where \(t\) is in seconds. Determine the magnitude of the particle's displacement from \(t=0\) to \(t=3 \mathrm{~s}\). Use Simpson's rule with \(n=100\) to evaluate the integrals. What is the magnitude of the particle's acceleration when \(t=2 \mathrm{~s} ?\)

A boat is traveling along a circular path having a radius of \(20 \mathrm{~m}\). Determine the magnitude of the boat's acceleration when the speed is \(v=5 \mathrm{~m} / \mathrm{s}\) and the rate of increase in the speed is \(\dot{v}=2 \mathrm{~m} / \mathrm{s}^{2}\).
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