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Chapter 2: Problem 21

A particle moving along the \(x\) -axis has its position described by the function \(x=\left(2 t^{2}-t+1\right) \mathrm{m},\) where \(t\) is in \(\mathrm{s}\). At \(t=2 \mathrm{s}\) what are the particle's (a) position, (b) velocity, and (c) acceleration?

Short Answer

Expert verified
At t = 2s, the particle's position is 5m, velocity is 7 m/s and acceleration is 4 m/s虏.

Step by step solution

01

Compute the Position

The position of the particle at any time \(t\) is given by \(x=2 t^{2}-t+1\). To find the position of the particle at \(t = 2s\), simply substitute \(t\) with 2 in the given equation, Hence, \(x = 2*(2^2) - 2 + 1 = 5m\). So, the particle is located 5m from the origin at t = 2 seconds.
02

Compute the Velocity

The velocity of the particle at any time \(t\) is the derivative of the position function. Differentiating the function \(x=2 t^{2}-t+1\) with respect to \(t\), we get \(v=dx/dt=4t-1\). Therefore, velocity at \(t = 2s\), is found by substituting \(t\) with 2 in the velocity equation, Hence, \(v = (4*2) - 1 = 7 m/s\). So, the velocity of the particle at \(t = 2s\) is 7 m/s.
03

Compute the Acceleration

Acceleration is the derivative of the velocity function or the second derivative of the position function. Differentiating the velocity function \(v=4t-1\) with respect to \(t\), we get \(a=dv/dt=4 m/s^2\). The acceleration does not depend on time (is constant), so the acceleration at any time instance, including \(t = 2s\), is \(4 m/s^2\).

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Key Concepts

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

Position Function
The **position function** tells us where an object is located along a certain axis at any moment in time. In the given exercise, the position of a particle moving along the x-axis is described by the equation \(x = 2t^2 - t + 1\), where \(t\) is the time in seconds, and \(x\) is the position in meters.
This equation is a quadratic polynomial, meaning it represents a parabolic trajectory when graphed. Each term in the equation contributes to the particle's motion over time:
  • The term \(2t^2\) indicates how the position changes quadratically with time, affecting the curvature of the path.
  • \(-t\) indicates a linear position change, contributing to the direction in which the path shifts.
  • The constant \(+1\) shifts the entire function upwards by 1 unit meter.
To find the particle's position at a specific time \(t = 2\) seconds, we substitute \(t\) with 2 in the equation:
  • Plugging in values: \(x = 2(2)^2 - 2 + 1 = 5\) meters.
  • This means the particle is 5 meters from the origin at this instant.
Velocity
**Velocity** describes how quickly the position of an object changes with time. Mathematically, it's the first derivative of the position function concerning time \(t\). By differentiating the position function \(x = 2t^2 - t + 1\), we determine the particle's velocity function:
  • The derivative \(\frac{dx}{dt} = 4t - 1\) gives us the velocity, which informs us of the speed and direction at any given \(t\).
Applying this to \(t = 2\) seconds:
  • Substitute the time: \(v = 4(2) - 1 = 7\) meters per second.
  • Thus, the particle is moving along the x-axis at 7 meters per second at \(t = 2\). Positive values indicate movement in the positive x direction.
  • Velocity being dependent on time in our formula implies changes in speed over different instances.
Acceleration
**Acceleration** measures how the velocity of an object changes with time, and it's the derivative of the velocity function. For constant acceleration, the rate of speed change remains undisturbed over time. By differentiating the already obtained velocity function \(v = 4t - 1\), we can compute the acceleration function:
  • The derivative \(\frac{dv}{dt} = 4\) is constant, indicating steady acceleration.
This constant acceleration of 4 meters per second squared implies that the particle's speed increases uniformly:
  • At any time \(t\), including \(t = 2\) seconds, the acceleration is \(4 \, m/s^2\).
  • This unchanging rate means that every second, the particle鈥檚 velocity increases by 4 meters per second.
  • In physical terms, a steady pushing force leads to uniform acceleration.
Understanding acceleration helps predict an object's future states of motion and is crucial for motion analysis in physics.

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Most popular questions from this chapter

A car starts at the origin and moves with velocity \(\vec{v}=\) \((10 \mathrm{m} / \mathrm{s},\) northeast). How far from the origin will the car be after traveling for 45 s?

a. What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? b. What fraction of \(g\) is this? c. How far has the car traveled when it reaches 60 mph? Give your answer both in SI units and in feet.

The position of a particle is given by the function \(x=\) \(\left(2 t^{3}-9 t^{2}+12\right) \mathrm{m},\) where \(t\) is in \(\mathrm{s}\) a. At what time or times is \(v_{x}=0 \mathrm{m} / \mathrm{s} ?\) b. What are the particle's position and its acceleration at this \(\operatorname{time}(s) ?\)

You are given the kinematic equation or equations that are used to solve a problem. For each of these, you are to: a. Write a realistic problem for which this is the correct equation(s). Be sure that the answer your problem requests is consistent with the equation(s) given. b. Draw the pictorial representation for your problem. c. Finish the solution of the problem. $$(10 \mathrm{m} / \mathrm{s})^{2}=v_{0 y}^{2}-2\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(10 \mathrm{m}-0 \mathrm{m})$$

Careful measurements have been made of Olympic sprinters in the 100 -meter dash. A quite realistic model is that the sprinter's velocity is given by $$v_{x}=a\left(1-e^{-b t}\right)$$ where \(t\) is in \(\mathrm{s}, v_{x}\) is in \(\mathrm{m} / \mathrm{s},\) and the constants \(a\) and \(b\) are characteristic of the sprinter. Sprinter Carl Lewis's run at the 1987 World Championships is modeled with \(a=11.81 \mathrm{m} / \mathrm{s}\) and \(b=0.6887 s^{-1}\) a. What was Lewis's acceleration at \(t=0 \mathrm{s}, 2.00 \mathrm{s},\) and \(4.00 \mathrm{s} ?\) b. Find an expression for the distance traveled at time \(t\) c. Your expression from part \(\mathbf{b}\) is a transcendental equation, meaning that you can't solve it for \(t .\) However, it's not hard to use rial and error to find the time needed to travel a specific distance. To the nearest \(0.01 \mathrm{s}\), find the time Lewis needed to sprint \(100.0 \mathrm{m} .\) His official time was \(0.01 \mathrm{s}\) more than your answer, showing that this model is very good, but not perfect.
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