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Chapter 12: Problem 3

A particle travels along a straight line with a velocity \(v=\left(12-3 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds. When \(t=1 \mathrm{~s}\), the particle is located \(10 \mathrm{~m}\) to the left of the origin. Determine the acceleration when \(t=4 \mathrm{~s}\), the displacement from \(t=0\) to \(t=10 \mathrm{~s}\), and the distance the particle travels during this time period.

Short Answer

Expert verified
The acceleration at \(t = 4\) s is \(-24\) m/s^2, the displacement from \(t = 0\) to \(t = 10\) s is \(-893\) m, and the total distance travelled from \(t = 0\) to \(t = 10\) s is \(871\) m.

Step by step solution

01

Find the acceleration function

Acceleration is the derivative of the velocity function. To find it, differentiate the given velocity function \(v = 12 - 3t^2\). Using standard rules of differentiation, the acceleration \(a = dv/dt = 0-2*3t = -6t \) m/s^2.
02

Substitute \(t=4\) s into the acceleration function

Substitution is used to find specific values of functions. To find acceleration at \(t=4\) s, substitute \(t=4\) into the acceleration function: \(a = -6 * 4 = -24\) m/s^2. So, the acceleration at \(t=4\) s is -24 m/s^2.
03

Find the displacement function

Displacement is the integral of the velocity function. To find it, integrate the given velocity function \(v = 12 - 3t^2\). Using standard rules of integration,with \(t=1\), the particles is at \(10m\) left of the origin. So, the displacement \(s = \int v dt = \int (12-3t^2) dt = 12t - t^3 + C\). To find the constant of integration \(C\), use the initial condition of the particle that at \(t=1\) s, the position is -10 m (because it is to the left of the origin): \(-10 = 12*1 - 1 + C, so C = -1 -12 = -13\). Therefore, the displacement function is \(s = 12t - t^3 -13\).
04

Calculate displacement from \(t=0\) to \(t=10\) s

Calculate the displacement by substituting the time values into displacement function. The displacement at \(t=10\) s is \(s = 12*10 - 10^3 - 13 = 120 - 1000 - 13 = -893\) m. The negative sign indicates that the particle is 893 m to the left of the origin at \(t=10\) s.
05

Calculate the total distance travelled

To calculate total distance travelled by the particle, it is significant to identify the intervals in which the velocity function changes its sign. Between these values of \(t\), the particle changes direction. The velocity function \(v=12 -3t^2\) equals zero for \(t= \sqrt[2]{4} = \pm 2\). Therefore, the intervals to consider are from \(t=0\) to \(t=2\) (where the particle moves in one direction), and from \(t=2\) to \(t=10\) (where the particle goes in the opposite direction). To obtain the distance, one must compute the absolute values of the displacements for these intervals, and then add them together. Following these steps, the total travelled distance results as \( | S_{[0,2]} | + | S_{[2,10]} | = | S_2 - S_0| + |S_{10} - S_2 | = | 24 - 8 - 13 | + | -893 - 24 + 8 + 13 | = |3| + | -868 | = 3 + 868 = 871\) m.

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Key Concepts

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

Understanding the Velocity Function
In the realm of particle motion, the velocity function is crucial. It allows us to describe how the speed of a particle changes over time. In our exercise, the velocity function is given as \( v = 12 - 3t^2 \). This function tells us the rate at which the particle moves along a straight line at any time \( t \).

Key characteristics of this function include:
  • It is a quadratic function in \( t \). The negative sign before \( 3t^2 \) indicates that the velocity decreases as time increases.
  • \( v \) becomes zero at a certain \( t \), which suggests that the particle stops momentarily before possibly reversing direction.
Understanding the velocity function allows us to predict the future position of the particle or analyze its past behavior. When graphed, a velocity function can show how fast the particle is moving at different points in time.
Decoding the Acceleration Function
Acceleration is the rate of change of velocity over time. In simple terms, it tells us how quickly the velocity itself is changing. To find the acceleration function from a velocity function, we differentiate the velocity function. Here, the velocity function \( v = 12 - 3t^2 \) is differentiated, giving us the acceleration function \( a = -6t \).

This calculation reveals important aspects about the particle's motion:
  • A linear function with a slope of \(-6\), implying acceleration is constant and negative over time, meaning the particle decelerates consistently.
  • At \( t = 4 \) seconds, substitution into the acceleration function yields \( a = -24 \ \text{m/s}^2 \), clearly indicating the rate of deceleration at that moment.
Acceleration provides important insights into the dynamics of the particle's motion, especially in analyzing changes as the particle speeds up or slows down.
Calculating Displacement
Displacement refers to the change in position of a particle. Unlike distance, which is always positive, displacement can be negative if the final position is left of the starting position.

To find displacement over a time interval, we integrate the velocity function. The given velocity leads to the displacement function \( s = 12t - t^3 - 13 \) by carefully integrating and finding the constant of integration. Using this function, evaluating between specific times gives the net change in position. From \( t = 0 \) to \( t = 10 \), substituting yields \(-893\ \text{m}\), meaning the particle is 893 meters left of its starting point at \( t = 0 \).

This mathematical process highlights how integration transforms velocity changes into a quantifiable positional change.
Integration of Velocity Function
Integration is the reverse of differentiation. In the context of motion, integrating a velocity function provides the position or displacement function.

The process involves:
  • Determining the indefinite integral of the velocity function.
  • Incorporating initial positional conditions to solve for any constants of integration, ensuring the displacement equation accurately reflects the particle's initial state.
For our problem, integrating \( v = 12 - 3t^2 \) resulted in \( s = 12t - t^3 + C \). Using initial conditions refined it to \( s = 12t - t^3 - 13 \). This integration provides a way to calculate the particle's positional shift over any given time period.
Finding Acceleration through Differentiation of Velocity
Differentiation involves finding how a quantity changes over time, essential in determining the acceleration from a velocity function. For our particle, differentiating the velocity \( v = 12 - 3t^2 \) with respect to time \( t \) is straightforward. Using basic calculus rules, the derivative of \( 12 - 3t^2 \) gives \( a = -6t \).

This process highlights:
  • How calculus smoothly transitions from analyzing speed changes to evaluating forces acting on the particle, as acceleration is directly linked to force.
  • The ease of determining instant accelerative behavior at any time \( t \) by simply substituting \( t \) into the derivative function.
Differentiation, therefore, is an invaluable tool in physics for understanding and predicting the acceleration of particles in motion.

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Most popular questions from this chapter

If a particle moves along a path such that \(r=\left(e^{a t}\right) \mathrm{m}\) and \(\theta=t\), where \(t\) is in seconds, plot the path \(r=f(\theta)\), and determine the particle's radial and transverse components of velocity and acceleration.

A car is traveling north along a straight road at \(50 \mathrm{~km} / \mathrm{h}\). An instrument in the car indicates that the wind is coming from the east. If the car's speed is \(80 \mathrm{~km} / \mathrm{h}\), the instrument indicates that the wind is coming from the northeast. Determine the speed and direction of the wind.

Ball \(A\) is thrown vertically upwards with a velocity of \(v_{0} .\) Ball \(B\) is thrown upwards from the same point with the same velocity \(t\) seconds later. Determine the elapsed time \(t<2 v_{0} / g\) from the instant ball \(A\) is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

The football player at \(A\) throws the ball in the \(y-z\) plane with a speed \(v_{A}=50 \mathrm{ft} / \mathrm{s}\) and an angle \(\theta_{A}=60^{\circ}\) with the horizontal. At the instant the ball is thrown, the player is at \(B\) and is running at a constant speed of \(v_{B}=23 \mathrm{ft} / \mathrm{s}\) along the line \(B C\). Determine if he can reach point \(C\), which has the same elevation as \(A\), before the ball gets there.

Starting from rest, the cable can be wound onto the drum of the motor at a rate of \(v_{A}=\left(3 t^{2}\right) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds. Determine the time needed to lift the load \(7 \mathrm{~m}\)
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