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Chapter 2: Problem 52

The acceleration of a bus is given by \(a_{x}(t)=\alpha t\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s},\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?\) (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(x-t\) graphs for the motion.

Short Answer

Expert verified
(a) 6.8 m/s at t = 2.0 s; (b) 11.8 m at t = 2.0 s; (c) Quadratic and cubic curves for velocity and position.

Step by step solution

01

Understand the given acceleration function

We are given that the acceleration function is \(a_x(t) = \alpha t\), where \(\alpha = 1.2 \, \mathrm{m/s^3}\). This tells us how the acceleration varies with time. We need to integrate this function to find velocity and position as functions of time.
02

Integrate acceleration to find velocity

The velocity function \(v_x(t)\) can be found by integrating the acceleration function. Integrate \(a_x(t) = 1.2t\) with respect to \(t\). The integral is: \[v_x(t) = \int 1.2t \, dt = \frac{1.2}{2}t^2 + C = 0.6t^2 + C\]where \(C\) is the constant of integration determined by initial conditions.
03

Use initial velocity to find constant C

At \(t = 1.0 \, \mathrm{s}\), the velocity \(v_x(1) = 5.0 \, \mathrm{m/s}\). Substitute these values into the velocity equation:\[5.0 = 0.6(1.0)^2 + C\]Solve for \(C\): \[5.0 = 0.6 + C \rightarrow C = 4.4\]Thus, \(v_x(t) = 0.6t^2 + 4.4\).
04

Calculate velocity at t = 2.0 s

Substitute \(t = 2.0\) into the velocity equation:\[v_x(2.0) = 0.6(2.0)^2 + 4.4 = 2.4 + 4.4 = 6.8 \, \mathrm{m/s}\]Thus, the velocity at \(t = 2.0 \, \mathrm{s}\) is \(6.8 \, \mathrm{m/s}\).
05

Integrate velocity to find position

To find the position function \(x(t)\), integrate the velocity function \(v_x(t) = 0.6t^2 + 4.4\):\[x(t) = \int (0.6t^2 + 4.4) \, dt = 0.2t^3 + 4.4t + C_2\]where \(C_2\) is another constant of integration.
06

Use initial position to find constant C2

At \(t = 1.0 \, \mathrm{s}\), the position \(x(1) = 6.0 \, \mathrm{m}\). Substitute these values into the position equation:\[6.0 = 0.2(1.0)^3 + 4.4(1.0) + C_2\]Solve for \(C_2\):\[6.0 = 0.2 + 4.4 + C_2 \rightarrow C_2 = 1.4\]Thus, \(x(t) = 0.2t^3 + 4.4t + 1.4\).
07

Calculate position at t = 2.0 s

Substitute \(t = 2.0\) into the position equation:\[x(2.0) = 0.2(2.0)^3 + 4.4(2.0) + 1.4 = 1.6 + 8.8 + 1.4 = 11.8 \, \mathrm{m}\]Thus, the position at \(t = 2.0 \, \mathrm{s}\) is \(11.8 \, \mathrm{m}\).
08

Sketch the a-t, v-t, and x-t graphs

The graph of \(a_x(t) = 1.2t\) is a straight line through the origin with slope 1.2. The graph of \(v_x(t) = 0.6t^2 + 4.4\) is a quadratic curve opening upwards. The graph of \(x(t) = 0.2t^3 + 4.4t + 1.4\) is a cubic curve. Plot these functions with respective variable on the y-axis and time \(t\) on the x-axis.

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly an object's velocity changes with time. In kinematics, acceleration is often provided as a function of time. In this exercise, the acceleration is given by the formula \( a_x(t) = \alpha t \), where \( \alpha = 1.2 \; \mathrm{m/s^3} \).
This means the acceleration increases linearly over time, making it a time-dependent acceleration.
This linear relationship is represented as a straight line on an acceleration-time graph, starting from the origin and having a slope of 1.2.
  • The slope indicates how steeply acceleration increases over time.
  • Positive slope means acceleration increases with time.
Understanding acceleration is crucial as it lays the foundation for calculating velocity and position through integration.
Velocity
Velocity is the rate at which an object's position changes over time and is itself dependent on acceleration. From acceleration \( a(t) = \alpha t \), we can find velocity by integrating the acceleration function.
Performing this integration gives us the velocity function: \( v_x(t) = 0.6t^2 + C \). The constant of integration \( C \) is determined using initial conditions, like the initial velocity provided.
  • The velocity at \( t = 1.0 \; \mathrm{s} \) is 5.0 m/s, which helps us solve for \( C \).
  • Once found, it allows us to determine the velocity at any other time, such as at \( t = 2.0 \; \mathrm{s} \).
By substituting into the velocity equation, we find \( v_x(2.0) = 6.8 \; \mathrm{m/s} \). Integrating acceleration to find velocity demonstrates how changes in motion evolve over time.
Position
Position describes the location of an object over time. With kinematics, you can determine position by integrating the velocity function.
From our earlier step, the velocity function is \( v_x(t) = 0.6t^2 + 4.4 \). Integrating the velocity function yields the position function: \( x(t) = 0.2t^3 + 4.4t + C_2 \).
  • The constant \( C_2 \) is found using initial position conditions.
  • Given that \( x(1.0) = 6.0 \; \mathrm{m} \), we calculate \( C_2 = 1.4 \).
This gives us the full position function \( x(t) = 0.2t^3 + 4.4t + 1.4 \).
Calculating at \( t=2.0 \; \mathrm{s} \), we find the position to be 11.8 m.
Position, therefore, accumulates the effect of velocity over time, just as velocity accumulates the effect of acceleration.
Integration
Integration in kinematics is the process used to derive velocity from acceleration and position from velocity. It works by summing up the infinitesimal changes over time.
The integral of the acceleration function \( a(t) \) gives you the velocity function, and similarly, integrating the velocity function provides the position function.
  • Integration introduces a constant of integration, \( C \), which is essential to adjust according to initial conditions, making each solution unique to its scenario.
  • Each integration step transforms the linear, quadratic, or cubic relationships shown in the kinematics equations.
This integral process allows us to track the motion from acceleration to velocity and then to position, illustrating how motion equations are interconnected.
Graphs of Motion
Graphs of motion visually depict relationships between acceleration, velocity, and position over time, enhancing our understanding of kinematic equations.
Each type of graph provides unique insights.
  • The acceleration-time graph is a straight line, indicating uniform increase over time.
  • The velocity-time graph is a parabola opening upward reflecting increasing velocity as a function of time squared.
  • The position-time graph is cubic, showing how position accumulates complexly over time.
Visualizing these graphs can help clarify the motion characteristics and the effect of acceleration and initial conditions on velocity and position.

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Most popular questions from this chapter

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) 50.0 m; (ii) 100.0 m; (iii) 200.0 m?

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end of the first 1.00 min, its speed is 1610 km/h. (a) What is the average acceleration (in m/s\(^2\)) of the rocket (i) during the first 8.00 s and (ii) between 8.00 s and the end of the first 1.00 min? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) during the first 8.00 s and (ii) during the interval from 8.00 s to 1.00 min?

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a 10-s interval. What are the magnitude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right. (a) At the beginning of the interval, the astronaut is moving toward the right along the \(x\)-axis at 15.0 m/s, and at the end of the interval she is moving toward the right at 5.0 m/s. (b) At the beginning she is moving toward the left at 5.0 m/s, and at the end she is moving toward the left at 15.0 m/s. (c) At the beginning she is moving toward the right at 15.0 m/s, and at the end she is moving toward the left at 15.0 m/s.

A small object moves along the \(x\) -axis with acceleration \(a_{x}(t)=-\left(0.0320 \mathrm{~m} / \mathrm{s}^{3}\right)(15.0 \mathrm{~s}-t) .\) At \(t=0\) the object is at \(x=-14.0 \mathrm{~m}\) and has velocity \(v_{0 x}=8.00 \mathrm{~m} / \mathrm{s} .\) What is the \(x\) -coordinate of the object when \(t=10.0 \mathrm{~s} ?\)

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 m away at an average speed of 3.00 m/s, returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?
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