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Chapter 2: Problem 46

An object moves in the \(+x\) -direction at a speed of \(40 \mathrm{~m} / \mathrm{s}\). As it passes through the origin, it starts to experience a constant acceleration of \(3.5 \mathrm{~m} / \mathrm{s}^{2}\) in the \(-x\) -direction. (a) What will happen next? (1) The object will reverse its direction of travel at the origin; (2) the object will keep traveling in the \(+x\) -direction; (3) the object will travel in the \(+x\) -direction and then reverses its direction. Why? (b) How much time elapses before the object returns to the origin? (c) What is the velocity of the object when it returns to the origin?

Short Answer

Expert verified
(3) The object reverses direction; Time: 22.9 s; Velocity: -40 m/s

Step by step solution

01

Understand the Motion Problem

The object is initially moving in the positive x-direction with a velocity of \(40 \mathrm{~m/s}\) and then experiences a negative acceleration \(a = -3.5 \mathrm{~m/s}^2\). We need to determine the behavior of the object over time, when it returns to the origin, and what its velocity will be at that point.
02

Determine the Future Motion

Given the negative acceleration, the object will first slow down, stop, and then reverse direction. Thus, the correct option is (3): the object will travel in the \(+x\) direction and then reverse its direction. This happens because the negative acceleration will eventually bring the velocity to zero and then make it negative, indicating a reversal in direction.
03

Calculate Time to Return to the Origin

We use the kinematic equation to find the time it takes for the object to return to the origin: \[ x = v_0 t + \frac{1}{2} a t^2\]Where \(x = 0\), \(v_0 = 40 \mathrm{~m/s}\), and \(a = -3.5 \mathrm{~m/s}^2\). Solving for the non-trivial solution: \[ 0 = 40t - \frac{1}{2} \times 3.5 \times t^2\]Rewriting and factoring, we get: \[ 0 = t(40 - 1.75t)\]Thus, the time \( t \) is given by: \[ t = \frac{40}{1.75} = 22.857 \approx 22.9 \mathrm{~s} \].
04

Calculate the Velocity at the Origin's Return

To find the velocity when returning, use the equation: \[ v = v_0 + a t \]Substituting given values:\[ v = 40 + (-3.5)(22.857) \approx 40 - 80 = -40 \mathrm{~m/s} \]This negative velocity confirms the object moves in the \(-x\) when it returns to the origin.

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Key Concepts

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

Kinematic Equations
Kinematic equations are crucial tools in understanding motion in one dimension, especially when dealing with constant acceleration. These equations relate parameters like displacement, velocity, time, and acceleration. In our exercise, the equation we used is: \[ x = v_0 t + \frac{1}{2} a t^2 \] which helps determine how far an object travels over time given its initial velocity \(v_0\) and acceleration \(a\).
  • \( x \) is the displacement from the origin, which in our case is 0 when the object returns.
  • \( v_0 \) is the initial velocity, here it's 40 m/s in the positive x-direction.
  • \( a \) is the constant acceleration, which is -3.5 m/s虏 in this problem.
  • \( t \) is the time, which is what we solved for to find when the object returns to the origin.
As you solve these equations, remember that signs are significant. They indicate direction, which is vital when determining outcomes of motion under constant acceleration.
Constant Acceleration
Constant acceleration means that an object's velocity changes at a consistent rate over time. This kind of acceleration, either speeding up or slowing down, is common in physics problems. In our exercise, the object undergoes a constant acceleration of -3.5 m/s虏. This negative value indicates it is opposite to the initial velocity, causing the object to eventually stop and reverse direction.The constant acceleration affects:
  • How fast an object can come to a stop or reverse direction.
  • The time needed for changes in motion, calculated using kinematic equations.
With constant acceleration:
  • The change in velocity (\( \Delta v \)) is linear over the time interval \( t \), so \( \Delta v = a \times t \).
  • This regular rate allows precise calculations for motion predictions, letting us use simple equations to solve for variables like time and velocity.
Velocity Reversal
Velocity reversal is when an object switches direction during its motion. Initially moving towards a specific direction, but due to an opposing force or acceleration, it eventually halts and begins moving the opposite way. In the context of our problem, the object starts traveling in the positive x-direction at 40 m/s but reverses due to a negative acceleration of -3.5 m/s虏.
  • First, the object decelerates until its velocity becomes zero. This is the point where it stops momentarily.
  • Then, the continuous negative acceleration pulls it towards the negative x-direction, causing a velocity reversal.
When calculating, it's helpful to:
  • Find the time at which the velocity is zero using the equation \( v = v_0 + a t \).
  • Understand that at any point when \( v \) becomes negative, it has reversed direction.
This reversal is confirmed when the velocity at the origin鈥檚 return is -40 m/s, indicating full reversal and movement in the opposite direction from where it started.

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Most popular questions from this chapter

A car initially at rest experiences loss of its parking brake and rolls down a straight hill with a constant acceleration of \(0.850 \mathrm{~m} / \mathrm{s}^{2},\) traveling a total of \(100 \mathrm{~m}\) Let's designate the first half of the distance as phase 1 with a subscript of 1 for those quantities, and the second half as phase 2 with a subscript of \(2 .\) (a) The car's speeds at the end of each phase should be related by which condition (1) \(v_{1}<\frac{1}{2} v_{2},\) (2) \(v_{1}=\frac{1}{2} v_{2},\) or (3) \(v_{1}>\frac{1}{2} v_{2} ?\) (b) Now calculate the two speeds and compare them quantitatively.

From street level, Superman spots Lois Lane in troublethe evil villain, Lex Luthor, is dropping her from near the top of the Empire State Building. At that very instant, the Man of Steel starts upward at a constant acceleration to attempt a midair rescue of Lois. Assuming she was dropped from a height of \(300 \mathrm{~m}\) and that Superman can accelerate straight upward at \(15.0 \mathrm{~m} / \mathrm{s}^{2},\) determine (a) how far Lois falls before he catches her, (b) how long Superman takes to reach her, and (c) their speeds at the instant he reaches her. Comment on whether these speeds might be a danger to Lois, who, being a mere mortal, might get hurt running into the impervious Man of Steel if the speeds are too great.

The time it takes for an object dropped from the top of cliff A to hit the water in the lake below is twice the time it takes for another object dropped from the top of cliff \(B\) to reach the lake. (a) The height of cliff \(A\) is (1) onehalf, (2) two times, (3) four times that of cliff B. (b) If it takes \(1.80 \mathrm{~s}\) for the object to fall from cliff \(\mathrm{A}\) to the water, what are the heights of cliffs \(\mathrm{A}\) and \(\mathrm{B} ?\)

A motorist travels \(80 \mathrm{~km}\) at \(100 \mathrm{~km} / \mathrm{h},\) and \(50 \mathrm{~km}\) at \(75 \mathrm{~km} / \mathrm{h}\). What is the average speed for the trip?

The driver of a pickup truck going \(100 \mathrm{~km} / \mathrm{h}\) applies the brakes, giving the truck a uniform deceleration of \(6.50 \mathrm{~m} / \mathrm{s}^{2}\) while it travels \(20.0 \mathrm{~m}\). (a) What is the speed of the truck in kilometers per hour at the end of this distance? (b) How much time has elapsed?
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