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The equation of motion for a person riding a bicycle is \(x=6.0 \mathrm{~m}+(4.5 \mathrm{~m} / \mathrm{s}) t\). (a) Where is the bike at \(t=2.0 \mathrm{~s}\) ? (b) At what time is the bike at the location \(x=24 \mathrm{~m}\) ?

Short Answer

Expert verified
(a) 15.0 m; (b) 4.0 s.

Step by step solution

01

Plug in the time for part (a)

To find the location of the bike at time \( t = 2.0 \, \text{s} \), we will substitute \( t = 2.0 \, \text{s} \) into the equation of motion \( x = 6.0 \, \text{m} + (4.5 \, \text{m/s}) \, t \).
02

Calculate location for part (a)

Substituting \( t = 2.0 \, \text{s} \) gives: \[x = 6.0 \, \text{m} + (4.5 \, \text{m/s}) \times 2.0 \, \text{s} = 6.0 \, \text{m} + 9.0 \, \text{m} = 15.0 \, \text{m}\]. So, at \( t = 2.0 \, \text{s} \), the bike is at \( 15.0 \, \text{m} \).
03

Set the equation for part (b)

For part (b), set the equation of motion equal to \( 24 \, \text{m} \) and solve for \( t \): \( 6.0 \, \text{m} + (4.5 \, \text{m/s}) \, t = 24 \, \text{m} \).
04

Solve for time in part (b)

Subtract \( 6.0 \, \text{m} \) from both sides to isolate the term with \( t \): \[(4.5 \, \text{m/s}) \, t = 24 \, \text{m} - 6.0 \, \text{m} = 18.0 \, \text{m}\].Divide both sides by \( 4.5 \, \text{m/s} \) to solve for \( t \): \[t = \frac{18.0 \, \text{m}}{4.5 \, \text{m/s}} = 4.0 \, \text{s}\]. Thus, the bike is at \( 24 \, \text{m} \) at \( t = 4.0 \, \text{s} \).

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Key Concepts

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

Linear Motion
Linear motion refers to the movement of an object in a straight line. When analyzing linear motion, we often look at how an object moves with respect to time along a straight path. This is a foundational concept in physics and helps us understand the motion of everyday objects, such as bicycles or cars, without the complexities of curves or angles.
In linear motion, there are a few essential elements to consider:
  • Displacement: This is the object's change in position from its starting point.
  • Velocity: This describes both the speed and direction of an object's movement.
  • Acceleration: In cases where the object's velocity changes, acceleration comes into play.
In the context of the given exercise, the position of the bicycle at any given time follows a straight path and can be described mathematically using the equation of motion.
Velocity
Velocity is a key concept when studying the motion of objects. It tells us both the speed and direction of an object's movement. While speed is a scalar quantity concerned only with how fast something is moving, velocity is a vector quantity that gives more information by including direction.
The equation used in the exercise, \( x = 6.0 \, \text{m} + (4.5 \, \text{m/s}) t \), reflects this idea. Here, the term \((4.5 \, \text{m/s})\) represents the constant velocity of the bicycle—in other words, it moves 4.5 meters every second in a particular direction.
Thus, velocity is essential in understanding how far and how quickly the bicycle is moving over a period of time. In this situation where velocity is constant, there is no need to consider any acceleration. The movement remains uniform, simplifying calculations.
Position-Time Relationship
The position-time relationship is an invaluable concept for describing how the position of an object changes over time. This relationship can be derived and expressed through the equation of motion, particularly useful for determining an object's location at a specific time or the time at which it will reach a certain position.
In our bicycle exercise, the position-time relationship is given by the equation \(x = 6.0 \, \text{m} + (4.5 \, \text{m/s}) t\). This indicates that the bicycle starts at a position of 6 meters and its position changes linearly with time because of its constant velocity of 4.5 meters per second.
For example, substituting different values of \(t\) can show various positions of the bike. This approach provides a clear, simple way to visualize and predict the bike's motion at any given time or reach any specified position. Understanding this relationship helps students solve practical problems and analyze real-world situations concerning motion.

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