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Chapter 2: Problem 20

Assume a canister in a straight tube moves with a constant acceleration of \(-4.00 \mathrm{~m} / \mathrm{s}^{2}\) and has a velocity of \(13.0 \mathrm{~m} / \mathrm{s}\) at \(\ell=0 .\) (a) What is its velocity at \(t=1.00 \mathrm{~s} ?\) (c) At \(t=2.50\) s? (b) At \(t=2.00 \mathrm{~s}^{2}\) ( (d) At \(t=4.00 \mathrm{s?}\) (e) Describe the shape of the canister's velocity versus time graph. (f) What two things must be known at a given time to predict the canister's velocity at any later time?

Short Answer

Expert verified
The velocities of the canister at \(t=1.00s, t=2.50s, t=2.00s\), and \(t=4.00s\) are calculated using the equation of motion. The velocity of canister gradually decreases due to negative acceleration. The graph representing velocity vs. time is a straight line sloping downwards. The initial velocity and the acceleration must be known to predict the canister's velocity.

Step by step solution

01

Calculate Velocity at \(t=1.00\) s

Use the first equation of motion \(v_f = v_i + a \cdot t\) where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration and \(t\) is the time. Substitute \(v_i = 13.0 \mathrm{~m/s}\), \(a = -4.00 \mathrm{~m/s^2}\), and \(t = 1.00 s\) to find \(v_f\).
02

Calculate Velocity at \(t=2.50\) s

Use the first equation of motion with \(v_i = 13.0 \mathrm{~m/s}\), \(a = -4.00 \mathrm{~m/s^2}\), and \(t = 2.50 s\) to find the new \(v_f\).
03

Calculate Velocity at \(t=2.00 s\)

Again, use the first equation of motion with \(t = 2.00 s\) to find the velocity at this time.
04

Calculate Velocity at \(t=4.00 s\)

Repeat prior steps for time \(t = 4.00 s\) to find the velocity at this moment.
05

Describe the Graph

As the acceleration is constant and negative, the velocity vs. time graph will be a straight line that decreases over time, starting from the initial velocity and sloping downwards from there.
06

Predicting Future Velocity

To predict the canister's velocity at any later time, we need to know both the initial velocity at a known time and the constant acceleration.

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Key Concepts

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

Velocity
Velocity is a fundamental concept in kinematics that describes the speed of an object in a specific direction. It is different from speed, which only measures how fast an object is moving without regard to direction. Velocity is a vector quantity, meaning it has both magnitude and direction.
  • Magnitude: How fast the object is moving.
  • Direction: The path the object is taking.
In the context of the canister moving in a straight tube, if the velocity at the initial point (\(t=0\)) is given as \(13.0 \,\mathrm{m/s}\), it indicates how fast and in which direction the canister is moving at that very moment. Remembering that velocity can change over time helps us understand how forces (like acceleration) influence motion.
Acceleration
Acceleration is the rate at which an object's velocity changes over time. It tells us how fast the velocity of an object is increasing or decreasing. Like velocity, acceleration is also a vector quantity, involving magnitude and direction.
  • Positive Acceleration: When speed increases in a particular direction.
  • Negative Acceleration (Deceleration): When speed decreases, like in the case of the canister, which has an acceleration of \(-4.00 \, \mathrm{m/s^2}\).
In this exercise, the constant negative acceleration indicates that the velocity of the canister is decreasing over time. This happens at a steady rate due to the acceleration being constant, resulting in uniform changes in velocity.
Equations of Motion
The equations of motion are mathematical formulas used to predict an object's future state in terms of velocity, position, and acceleration. These equations are crucial in kinematics for calculating how different variables are interrelated over specific time intervals.
  • First Equation of Motion: \(v_f = v_i + a \cdot t\)
This equation helps us compute final velocity \(v_f\) when initial velocity \(v_i\), acceleration \(a\), and time \(t\) are known. In the exercise context, using given values, substituting them into the first equation of motion allows for calculating precise future velocities. This clear methodology is powerful, enabling predictive insights on an object's movement.
Velocity-Time Graph
A velocity-time graph visually represents how an object's velocity changes over time. We'll draw it with the initial velocity on the y-axis at time zero and observe how it progresses.
  • With constant negative acceleration, as present in this exercise with \(-4.00 \, \mathrm{m/s^2}\), the graph appears as a straight, downward-sloping line.
  • The slope of this line equals the acceleration.
By examining the graph, it's possible to identify key patterns in motion, such as when the object is speeding up, slowing down, or moving at a constant velocity. Understanding the shape and slope provides clear insights into motion dynamics, making it easier to predict future states based on current information.

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Most popular questions from this chapter

To qualify for the finals in a racing event, a race car must achieve an average speed of \(250 \mathrm{~km} / \mathrm{h}\) on a track with a total length of \(1600 \mathrm{~m}\). If a particular car covers the first half of the track at an average speed of \(230 \mathrm{~km} / \mathrm{h}\), what minimum average speed must it have in the second half of the event in order to qualify?

A ball is thrown vertically upward with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started?

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at \(+40 \mathrm{ft} / \mathrm{s}^{2}\). After some time \(t_{1}\), the rocket engine is shut down and the sled moves with constant velocity \(v\) for a time \(l_{2}\). If the total distance traveled by the sled is \(17500 \mathrm{ft}\) and the total time is \(90 \mathrm{~s}\), find (a) the times \(t_{1}\) and \(t_{2}\) and (b) the velocity \(\nu\). At the \(17500-\) ft mark, the sled begins to accelerate at \(-20 \mathrm{ft} / \mathrm{s}^{2}\), (c) What is the final position of the sled when it comes to rest? (d) How long does it take to come to rest?

A truck on a straight road starts from rest and accelerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) until it reaches a speed of \(20 \mathrm{~m} / \mathrm{s}\). Then the truck travels for \(20 \mathrm{~s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.0 \mathrm{~s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck during the motion described?

Two students are on a balcony \(19.6 \mathrm{~m}\) above the street. One student throws a ball vertically downward at \(14.7 \mathrm{~m} / \mathrm{s} ;\) at the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (a) What is the difference in the two balls' time in the air? (b) What is the velocity of each ball as it strikes the ground? (c) How far apart are the balls \(0.800 \mathrm{~s}\) after they are thrown?
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