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Chapter 2: Problem 5

The acceleration of a particle is given by \(a=2 t-10\) where \(a\) is in meters per second squared and \(t\) is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at \(t=0\) is \(s_{0}=-4 \mathrm{m},\) and the initial velocity is \(v_{0}=3 \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
Velocity: \( v(t) = t^2 - 10t + 3 \); Displacement: \( s(t) = \frac{t^3}{3} - 5t^2 + 3t - 4 \).

Step by step solution

01

Integrate Acceleration to Find Velocity

The acceleration is a function of time given by \( a(t) = 2t - 10 \). To find the velocity as a function of time, we need to integrate the acceleration with respect to time.\[ v(t) = \int (2t - 10)\, dt \]Carrying out the integration, we obtain:\[ v(t) = \int 2t\, dt - \int 10\, dt = t^2 - 10t + C\]Where \( C \) is the constant of integration.
02

Use Initial Condition to Solve for Constant in Velocity Function

We know the initial velocity at \( t=0 \) is \( v(0) = 3 \text{ m/s} \). Using this, we substitute into the velocity equation:\[ v(0) = 0^2 - 10 \cdot 0 + C = 3\]Thus, \( C = 3 \). Now, the velocity function is:\[ v(t) = t^2 - 10t + 3\]
03

Integrate Velocity to Find Displacement

With \( v(t) = t^2 - 10t + 3 \), we now find displacement as a function of time by integrating the velocity:\[ s(t) = \int (t^2 - 10t + 3)\, dt\]This results in:\[ s(t) = \frac{t^3}{3} - 5t^2 + 3t + C_2\]Where \( C_2 \) is another constant of integration.
04

Use Initial Displacement to Solve for Constant in Displacement Function

The initial displacement at \( t=0 \) is \( s(0) = -4 \text{ m} \). To find \( C_2 \), substitute \( t=0 \) into the displacement equation:\[s(0) = \frac{0^3}{3} - 5 \cdot 0^2 + 3 \cdot 0 + C_2 = -4\]So, \( C_2 = -4 \), giving us the displacement function:\[ s(t) = \frac{t^3}{3} - 5t^2 + 3t - 4\]

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Key Concepts

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

Dynamics
Dynamics is a branch of physics that deals with forces and their effects on motion. It focuses on how physical bodies interact with their surroundings. One key element in dynamics is understanding the effect of acceleration on the motion of an object.
The interaction often comes from forces, following principles laid out by Newton's laws of motion. These laws help us predict how an object will move when subjected to certain forces.
  • Newton's First Law: An object at rest stays at rest, and an object in motion stays in motion unless acted on by a net external force.
  • Newton's Second Law: The acceleration of an object is dependent on two variables - the net force acting upon the object and the mass of the object. It's often summarized by the equation: \[ F = m imes a \]
  • Newton's Third Law: For every action, there is an equal and opposite reaction.
Understanding aspects of dynamics is crucial when analyzing particle motion, as in this exercise, where we derive the velocity and displacement of a particle from its given acceleration.
Motion Equations
Motion equations are mathematical formulations that describe the relationships between displacement, velocity, acceleration, and time. They are fundamental in physics, especially when analyzing motion in dynamics. In this exercise, we utilized motion equations through the process of integration.
When we integrate acceleration, we obtain the velocity function, which gives the rate of change of position over time. Likewise, integrating velocity yields the displacement function, which tells us the position of the particle over time.
  • Velocity (\[ v(t) \]) formula derived: \[ v(t) = t^2 - 10t + 3 \]
  • Displacement (\[ s(t) \]) formula derived: \[ s(t) = \frac{t^3}{3} - 5t^2 + 3t - 4 \]
Effectively applying these equations enables us to solve complex dynamics problems by understanding how one quantity influences another.
Acceleration
Acceleration is the rate at which an object changes its velocity. It is a vector quantity, meaning it has both magnitude and direction. In this exercise, acceleration was given as a function of time, \[ a(t) = 2t - 10 \], illustrating how it changes over time.
Here's why acceleration is crucial:
  • It directly influences velocity; as it changes, the velocity changes.
  • Positive acceleration means the object is speeding up; negative means it's slowing down.
  • Uniform acceleration implies constant acceleration, simplifying calculations and predictions.
By understanding acceleration's role, one can predict how quickly an object will change its speed and in what direction, helping solve the integral calculus involved in motion problems.
Initial Conditions
Initial conditions are the specific values of variables at the start of the observation or calculation, playing an essential role in solving dynamic equations. They allow us to determine the constants of integration, bringing specificity to the derived functions.
In this exercise, the initial conditions provided were:
  • Initial velocity: \[ v_0 = 3 \,\text{m/s} \] ; helps determine the constant of integration in the velocity equation.
  • Initial displacement: \[ s_0 = -4 \,\text{m} \]; used to find the constant in the displacement equation.
By substituting these initial values into the integrated equations, we found the complete expressions for both velocity and displacement as functions of time. This process is essential in dynamics to accurately describe the motion of particles and objects.

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Most popular questions from this chapter

A particle in an experimental apparatus has a velocity given by \(v=k \sqrt{s},\) where \(v\) is in millimeters per second, the position \(s\) is millimeters, and the constant \(k=0.2 \mathrm{mm}^{1 / 2} \mathrm{s}^{-1}\). If the particle has a velocity \(v_{0}=3 \mathrm{mm} / \mathrm{s}\) at \(t=0,\) determine the particle position, velocity, and acceleration as functions of time, and compute the time, position, and acceleration of the particle when the velocity reaches \(15 \mathrm{mm} / \mathrm{s}\)

In a test of vertical leaping ability, a basketball player (a) crouches just before jumping, (b) has given his mass center \(G\) a vertical velocity \(v_{0}\) at the instant his feet leave the surface, and \((c)\) reaches the maximum height. If the player can raise his mass center 3 feet as shown, estimate the initial velocity \(v_{0}\) of his mass center in position ( \(b\) ).

For the baseball of Prob. \(2 / 45\) thrown upward with an initial speed of \(30 \mathrm{m} / \mathrm{s}\), determine the time \(t_{u}\) from ground to apex and the time \(t_{d}\) from apex to ground.

An amusement ride called the "corkscrew" takes the passengers through the upside-down curve of a horizontal cylindrical helix. The velocity of the cars as they pass position \(A\) is \(15 \mathrm{m} / \mathrm{s}\), and the component of their acceleration measured along the tangent to the path is \(g \cos \gamma\) at this point. The effective radius of the cylindrical helix is \(5 \mathrm{m},\) and the helix angle is \(\gamma=40^{\circ} .\) Compute the magnitude of the acceleration of the passengers as they pass position \(A\)

Train \(A\) is traveling at a constant speed \(v_{A}=\) \(35 \mathrm{mi} / \mathrm{hr}\) while car \(B\) travels in a straight line along the road as shown at a constant speed \(v_{B}\). A conductor \(C\) in the train begins to walk to the rear of the train car at a constant speed of \(4 \mathrm{ft} / \mathrm{sec}\) relative to the train. If the conductor perceives car \(B\) to move directly westward at \(16 \mathrm{ft} / \mathrm{sec},\) how fast is the car traveling?
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