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Chapter 12: Problem 40

The \(s-t\) graph for a train has been experimentally determined. From the data, construct the \(v-t\) and \(a-t\) graphs for the motion; \(0 \leq t \leq 40 \mathrm{~s}\). For \(0 \leq t \leq 30 \mathrm{~s}\), the curve is \(s=\left(0.4 t^{2}\right) \mathrm{m},\) and then it becomes straight for \(t \geq 30 \mathrm{~s}\).

Short Answer

Expert verified
The velocity-time graph for the motion of the train is \(v=0.8t\) when \(0 鈮� t 鈮� 30s\) and a constant when \(t \geq 30s\). The acceleration-time graph is \(a= 0.8\) when \(0 鈮� t 鈮� 30s\) and zero when \(t \geq 30s\).

Step by step solution

01

Derive the Velocity

Given the displacement-time equation for 0 鈮� t 鈮� 30 s as \(s=(0.4 t^{2})\), the velocity can be obtained by taking the derivative of the displacement equation with respect to time (t), i.e, \(v=0.4*2*t = 0.8t\), which is a straight line equation.
02

Derive Acceleration

The acceleration equation can be deduced by taking the derivative of the velocity equation with respect to time (t), i.e, \(a= 0.8\). This indicates that acceleration is constant when \(0 鈮� t 鈮� 30s\).
03

Velocity & Acceleration for \(t>30s\)

Since the displacement-time graph is a straight line (linear) when \(t \geq 30s\), its gradient will be constant so velocity will be a constant value and as rate of change of velocity which is acceleration is zero for \(t \geq 30s\).

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Key Concepts

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

Velocity-Time Graph
The velocity-time graph gives us valuable information about how a body's speed changes over time. It is derived from the displacement-time graph by understanding that velocity is the rate at which displacement changes. From the solution provided, we learned that for the time interval 0 鈮� t 鈮� 30 seconds, the displacement-time relationship is given by the equation \(s=0.4 t^2\).
Taking the derivative of \(s\) with respect to time \(t\) gives us the velocity \(v\), which is \(v = 0.8t\). This linear equation implies a constant change in velocity, forming a straight line on the velocity-time graph with a slope of 0.8. This tells us that the object speeds up at a constant rate over this period.
When \(t \geq 30 \) seconds, the displacement-time graph becomes a straight line, indicating no change in velocity. Therefore, the velocity remains constant beyond 30 seconds, resulting in a horizontal line on the velocity-time graph during this phase.
  • For 0 鈮� t 鈮� 30 s: A line with slope 0.8.
  • For t 鈮� 30 s: A horizontal line indicating constant velocity.
Acceleration-Time Graph
Acceleration describes how velocity changes over time and is depicted in an acceleration-time graph. To find acceleration, we take the derivative of the velocity equation. For the motion described in the exercise, the velocity equation for 0 鈮� t 鈮� 30 s is \(v = 0.8t\).
By taking the derivative of velocity with respect to time, we find the acceleration \(a\), which is a constant 0.8 m/s虏. This constant value indicates that between 0 and 30 seconds, the object experiences uniform acceleration. This would appear as a horizontal line on the acceleration-time graph at 0.8 m/s虏 for that time period.
For t 鈮� 30 s, since the velocity is constant, the acceleration drops to 0 because there is no change in velocity. Thus, the acceleration-time graph will show a horizontal line at 0 for t 鈮� 30 seconds.
  • For 0 鈮� t 鈮� 30 s: Horizontal line at 0.8 m/s虏.
  • For t 鈮� 30 s: Horizontal line at 0 m/s虏.
Displacement-Time Graph
The displacement-time graph provides insights into how the position of a moving object changes over time. In the context of the exercise, we start with the equation \(s = 0.4 t^2\) for the first 30 seconds. This indicates that the displacement increases quadratically over time, forming a parabolic curve. As time progresses, the object's position changes more rapidly due to increasing velocity.
For t 鈮� 30 seconds, the displacement continues to grow but at a constant rate, meaning the graph is a straight line from this point onward. This shows that the object moves with constant velocity due to the absence of acceleration.
  • For 0 鈮� t 鈮� 30 s: A parabolic curve indicating increasing speed.
  • For t 鈮� 30 s: A straight line showing constant speed and increasing displacement.
Understanding these graphs helps us visualize and comprehend the motion characteristics of an object in a straightforward way.

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Most popular questions from this chapter

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 partial surface of the cam is that of a logarithmic spiral \(r=\left(40 e^{0.05 \theta}\right) \mathrm{mm},\) where \(\theta\) is in radians. If the cam is rotating at a constant angular rate of \(\dot{\theta}=4 \mathrm{rad} / \mathrm{s}\), determine the magnitudes of the velocity and acceleration of the follower rod at the instant \(\theta=30^{\circ}\).

A particle is moving along a circular path having a 400-mm radius. Its position as a function of time is given by \(\theta=\left(2 t^{2}\right) \mathrm{rad},\) where \(t\) is in seconds. Determine the magnitude of the particle's acceleration when \(\theta=30^{\circ} .\) The particle starts from rest when \(\theta=0^{\circ} .\)

The position of a particle is defined by \(r=\\{5(\cos 2 t) \mathbf{i}+4(\sin 2 t) \mathbf{j}\\} \mathrm{m},\) where \(t\) is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particle when \(t=1 \mathrm{~s}\). Also, prove that the path of the particle is elliptical.

When a particle falls through the air, its initial acceleration \(a=g\) diminishes until it is zero, and thereafter it falls at a constant or terminal velocity \(v_{f} .\) If this variation of the acceleration can be expressed as \(a=\left(g / v_{f}^{2}\right)\left(v_{f}^{2}-v^{2}\right),\) determine the time needed for the velocity to become \(v=v_{f} / 2\). Initially the particle falls from rest.
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