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Chapter 14: Problem 15

A particle starts with an initial velocity of \(8 \mathrm{~m} / \mathrm{s}\) and moves along a straight line. It's acceleration time \(t\) after start is given by the expression \(A-B t\) where \(A\) and \(B\) are constants. stops. Hint: \(\left[\begin{array}{ll}t=0, & v=8 \mathrm{~m} / \mathrm{s} \\\ t=0 & x=0 \\ t=5 s & x=40 \\ t=5 s & v=0\end{array}\right] \quad\left[x=8 t+1.6 t^{2}-0.32 t^{3}\right]\)

Short Answer

Expert verified
In conclusion, the acceleration expression for the particle is \(a(t)=-\dfrac{8}{5}\), and the position expression is \(x(t) = 8t - \dfrac{4}{5}t^2\).

Step by step solution

01

Use the given data

The first hint states that at time t = 0, the velocity is 8 m/s and at t = 5 s, the velocity is 0. This data can be used to determine the constants A and B in the acceleration expression.
02

Calculate acceleration at t = 5 s

Since at t = 5 s, the velocity is 0, we can find the acceleration at this time. Using the initial velocity and final velocity, we can find the acceleration using the formula: \(v = u + at\) \(0 = 8 + a(5)\) Solving for the acceleration, we get: \(a = -\dfrac{8}{5} \mathrm{ms^{-2}}\)
03

Find the expression for acceleration

Now we know the acceleration at t = 5 s is -8/5 ms^-2. We can substitute this into the given acceleration expression: \(-\dfrac{8}{5} = A - B(5)\) We also know the acceleration expression A - Bt is valid for any time t. So, at t = 0, u = 8 m/s and a = A - B(0), acceleration is equal to A. Now, we have: \(A = -\dfrac{8}{5} \mathrm{ms^{-2}}\)
04

Find the constant B

Now that we found A, we can find B using the equation from Step 3: \(-\dfrac{8}{5} = A - B(5)\) Substituting A: \(B(5) = 0\) The constant B is 0. Now we have found the acceleration expression is: \(a(t)=-\dfrac{8}{5}\)
05

Integrate the acceleration expression to find the velocity expression.

We have the acceleration expression, \(a(t)=-\dfrac{8}{5}\), and we know that at t = 0, the velocity is 8 m/s. Integrate the acceleration expression with respect to time: \(v(t)=\int a(t) \, dt = 8 -\dfrac{8}{5}t\)
06

Integrate the velocity expression to find the position expression.

Integrate the velocity expression, \(v(t) = 8 -\dfrac{8}{5}t\), with respect to time: \(x(t)=\int v(t) \, dt = 8t - \dfrac{4}{5}t^2 + C\) Use the hint that at t = 0, the position is 0, to find the value for the constant C: \(0 = 8(0) - \dfrac{4}{5}(0)^2 + C\) C = 0. So the final position expression is: \(x(t) = 8t - \dfrac{4}{5}t^2\). In conclusion, the acceleration expression is \(a(t)=-\dfrac{8}{5}\), and the position expression is \(x(t) = 8t - \dfrac{4}{5}t^2\).

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Key Concepts

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

Acceleration
Acceleration is the rate of change of velocity over time. It tells us how quickly an object speeds up or slows down. In this exercise, we are given an acceleration expression in terms of constants, represented as \(A - B t\). The exercise states that at time \(t = 5 \,s\), the velocity reaches zero, indicating that the particle comes to a stop. By using this information, along with the initial conditions, we can solve for the constants \(A\) and \(B\), ultimately determining the specific acceleration behavior of our particle. Key points to remember about acceleration:
  • Positive acceleration means an increase in velocity.
  • Negative acceleration indicates a decrease in velocity (also known as deceleration).
  • Constant acceleration results in a uniform change in velocity over time.
Velocity
Velocity describes the speed of an object in a specific direction. It鈥檚 a vector quantity, unlike speed which is scalar. In this scenario, the initial velocity is given as \(8 \, ext{m/s}\). Upon performing integration on the acceleration function, the velocity function of the object is derived as \(v(t) = 8 - \frac{8}{5} t\). This equation helps us understand how the velocity varies over time. To break it down:
  • Initial velocity is \(8 \, ext{m/s}\), reducing linearly with time due to negative acceleration.
  • When \(t = 5 \, s\), the velocity is calculated to be zero, implying that the particle stops moving temporarily.
  • This equation shows how velocity and acceleration are closely related.
Position
Position refers to an object's location at a specific point in time. It's essential to track the particle's journey as it moves along a straight line. In this problem, we derive the position function through the integration of the velocity function, resulting in \(x(t) = 8t - \frac{4}{5}t^2\). This expression shows us:
  • At \(t = 0\), the position is \(0\), which helps us determine any constants in the integration.
  • The particle moves forward initially as \(8 t\) dominates initially, but decelerates as the \(- \frac{4}{5} t^2\) term catches up.
  • Overall position gives us insight on how far the particle has traveled over a period.
Integration
Integration is a mathematical method required to go from acceleration to velocity and from velocity to position. It is the process of finding the integral of a function over time, which comes quite handy in kinematics:
  • To find the velocity from acceleration, integrate the acceleration function: \[ v(t) = \int a(t) \, dt \]
  • To find the position from velocity, integrate the velocity function: \[ x(t) = \int v(t) \, dt \]
  • Each integration step introduces a constant of integration, which can be determined using initial conditions provided in the problem.
Integration allows us to determine how velocity and position evolve as a function of time, given the initial state of the particle.
Dynamics
Dynamics in physics examines the forces and motions of physical objects. This exercise is an application of dynamics because we evaluate and predict the motion of a particle based on initial conditions and changes in its motion over time. Crucially:
  • We are dealing with the motion of a particle along a straight line, influenced by its initial velocity and a given time-varying acceleration.
  • Through dynamics, we connect initial conditions to final states using mathematical descriptions (equations derived from integration).
  • Understanding dynamics allows us to predict how particles behave under various forces, leading to real-world applications like vehicle design and motion prediction.
Overall, this exercise highlights the necessity of dynamics in understanding motion in the realm of physics.

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Most popular questions from this chapter

A motorist takes 10 seconds to cover a distance of \(20 \mathrm{~m}\) and 15 seconds to cover a distance of \(40 \mathrm{~m} .\) Find the uniform acceleration of the car and the velocity at the end of 15 seconds. \(\left[0.267 \mathrm{~m} / \mathrm{s}^{2}, 4.7 \mathrm{~m} / \mathrm{s}\right]\)

A ball is thrown vertically up with a velocity of \(30 \mathrm{~m} / \mathrm{s}\). Determine the time (a) when th will be \(20 \mathrm{~m}\) above the point of projection, \((b)\) when its velocity will be \(5 \mathrm{~m} / \mathrm{s},(c)\) when \(\mathrm{t}_{\text {ball }}\) return back to its initial position. \([0.7 \mathrm{~g}, 2.55\)

0\. \(3.87 \mathrm{~cm}]\) The acceleration of a particle moving in a straight line is given by the relation \(a \sqrt{v}\). Its displacement and velocity at time \(t=2 \mathrm{~s}\) are \(42 \frac{2}{3} \mathrm{~m}\) and \(16 \mathrm{~m} / \mathrm{s}\) respectively. Find the displacement, velocity and acceleration of the particle at time \(t=3\) seconds.

A particle falls vertically in a medium whose resistance is proportional to the velocity of the particle. Find the velocity and distance travelled by the particle after a time \(t\). Hint: \(\left(\begin{array}{l}\text { Resistance, } R=k v^{2} \\ m a=m g-k v^{2}\end{array}\right) \quad\left(\begin{array}{l}v=\frac{g}{k}\left(1-e^{-k t}\right) \\ x=\left(\frac{g}{k} t+\frac{g}{k^{2}} e^{-k t}-\frac{g}{k^{2}}\right)\end{array}\right)\)

Two blocks \(P\) and \(Q\) are connected by a string of length \(12 \mathrm{~m}\) passing over a very small pulley. The arrangement is such that \(P\) moves horizontally and \(Q\) vertically up. When \(x=4 \mathrm{~m}\) it is found that velocity of \(P\) is \(3 \mathrm{~m} / \mathrm{s}\) towards right. Find the velocity of block. [Hint: Length of the string is constant, \(l=\) \(12 \mathrm{~m}\). Expressing displacements \(x, y\) of \(P\) and \(Q\) from same fixed point in terms of \(l\). $$ \begin{aligned} l &=(6-y)+A B \\ A B^{2} &=x^{2}+6^{2} \\ l &=(6-y)+\sqrt{x^{2}+6^{2}} \end{aligned} $$ This displacement relationship between \(P\) and \(Q\) can now be differentiated to get velocity and acceleration relationships] $$ \left[v_{q}=1.66 \mathrm{~m} / \mathrm{s}, y=1.21 \mathrm{~m}\right] $$
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