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Chapter 2: Problem 1

The velocity of a particle is given by \(v=25 t^{2}-80 t-\) \(200,\) where \(v\) is in feet per second and \(t\) is in seconds. Plot the velocity \(v\) and acceleration \(a\) versus time for the first 6 seconds of motion and evaluate the velocity when \(a\) is zero.

Short Answer

Expert verified
Velocity when acceleration is zero is \(v(1.6) = -264\) feet/second.

Step by step solution

01

Find the Expression for Acceleration

Acceleration is the derivative of velocity with respect to time. Given that the velocity is \(v = 25t^2 - 80t - 200\), the acceleration \(a(t)\) is found by differentiating this expression with respect to \(t\).
02

Differentiate to Find Acceleration Function

Differentiate \(v = 25t^2 - 80t - 200\) with respect to \(t\) to find the acceleration. \[ a(t) = \frac{d}{dt}(25t^2 - 80t - 200) = 50t - 80 \]
03

Plot Velocity and Acceleration

To plot the given functions, use a range of \(t\) values from 0 to 6 seconds. For each \(t\), calculate \(v(t)\) and \(a(t)\) using the expressions: \(v(t) = 25t^2 - 80t - 200\) and \(a(t) = 50t - 80\). Create a graph with time \(t\) on the x-axis and both \(v\) and \(a\) on the y-axis.
04

Find Time When Acceleration is Zero

Set the acceleration equal to zero to find the time at which it occurs. Solve the equation \(50t - 80 = 0\): \(50t = 80\) which simplifies to \(t = \frac{80}{50} = 1.6\) seconds.
05

Evaluate Velocity at Zero Acceleration

Substitute \(t = 1.6\) seconds into the velocity equation: \(v(1.6) = 25(1.6)^2 - 80(1.6) - 200\). Calculate to find the velocity at \(t = 1.6\).

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Key Concepts

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

Understanding Velocity and Acceleration
Velocity and acceleration are fundamental concepts in dynamics, which is a branch of physics that deals with the motion of objects. **Velocity** refers to the speed of an object in a given direction, while **acceleration** is the rate of change of velocity.

In the given exercise, the velocity of a particle is described by the equation \(v = 25t^2 - 80t - 200\). Here, \(v\) represents velocity in feet per second, and \(t\) is time in seconds. To find how fast an object's velocity is changing over time, we need to compute its acceleration.

The acceleration is obtained by differentiating the velocity equation with respect to time, resulting in the formula \(a(t) = 50t - 80\). This derivative tells you how much the velocity changes with respect to each second passing.
Exploring Derivatives in Physics
Derivation is a key mathematical tool used in physics to find rates of change. The process of differentiation helps to understand how a quantity varies with respect to another. In our problem, we use differentiation to find acceleration by examining the derivative of the velocity function.

Let's look at the given velocity function \(v = 25t^2 - 80t - 200\). By taking the derivative of this function, we transform it into an equation that represents acceleration. This derivative reflects how each component of the velocity equation contributes to the overall acceleration.
  • The term \(25t^2\) becomes \(50t\) when differentiated.
  • The term \(-80t\) becomes \(-80\).
  • The constant \(-200\) disappears as the derivative of a constant is zero.
Thus, we arrive at the acceleration function \(a(t) = 50t - 80\), which provides insights into the nature of an object's motion with increasing time.
Plotting Graphs for Enhanced Understanding
Graphical representations are powerful for visualizing relationships between different quantities in physics. They offer a clear picture of how variables like velocity and acceleration change over time.
To visualize the motion, plot the velocity \(v(t) = 25t^2 - 80t - 200\) and acceleration \(a(t) = 50t - 80\) over the first 6 seconds. Make sure to place time \(t\) on the x-axis and measure both \(v\) and \(a\) on the y-axis.
  • Velocity graph: Demonstrates how fast and in which direction the object is moving at a particular time.
  • Acceleration graph: Shows the rate of change of velocity, which can indicate whether the object is speeding up or slowing down.
Identifying when acceleration is zero, such as at \(t = 1.6\) seconds, helps to pinpoint specific moments in the motion when the object's velocity might momentarily stop changing. By substituting this time back into the velocity equation, you can find out what the velocity is when the motion is no longer accelerating.

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Most popular questions from this chapter

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The position \(s\) of a particle along a straight line is given by \(s=8 e^{-0.4 t}-6 t+t^{2},\) where \(s\) is in meters and \(t\) is the time in seconds. Determine the velocity \(v\) when the acceleration is \(3 \mathrm{m} / \mathrm{s}^{2}\)
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