/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 An automobile and a truck start ... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 69

An automobile and a truck start from rest at the same instant, with the automobile initially at some distance behind the truck. The truck has a constant acceleration of \(2.10 \mathrm{m} / \mathrm{s}^{2},\) and the antomobile an acceleration of 3.40 \(\mathrm{m} / \mathrm{s}^{2} .\) The automobile overtakes the truck after the truck has moved 40.0 \(\mathrm{m}\) . (a) How much time does it take the automobile to overtake the truck? (b) How far was the automobile behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x=0\) at the initial location of the truck.

Short Answer

Expert verified
(a) 6.17 s, (b) 24.88 m behind, (c) Truck: 12.96 m/s, Automobile: 20.98 m/s.

Step by step solution

01

Determine the time it takes for the truck to travel 40.0 m

Using the kinematic equation for constant acceleration, the distance covered by the truck is given by \( s_t = \frac{1}{2} a_t t^2 \), where \( s_t = 40.0 \text{ m} \) and \( a_t = 2.10 \text{ m/s}^2 \). Set up the equation: \( 40.0 = \frac{1}{2} \times 2.10 \times t^2 \). Solving for \( t \), we find \( t^2 = \frac{40.0}{1.05} = 38.1 \), hence \( t \approx 6.17 \text{ s} \).
02

Determine the position equation for the automobile

For the automobile with acceleration \( a_a = 3.40 \text{ m/s}^2 \), the displacement as a function of time is \( s_a = -s_0 + \frac{1}{2} a_a t^2 \), where \( s_0 \) is the initial distance behind the truck. At time \( t \), the automobile overtakes the truck, so \( s_a = 40.0 \text{ m} \).
03

Solve for the initial position of the automobile

Substitute \( t = 6.17 \text{ s} \) into the equation for the automobile: \( 40.0 \text{ m} = -s_0 + \frac{1}{2} \times 3.40 \times (6.17^2) \). Simplifying, \( 40.0 = -s_0 + 64.88 \), leading to \( s_0 = 64.88 - 40.0 = 24.88 \text{ m} \).
04

Determine the speed of each vehicle when they meet

Use the formula \( v = u + a t \), where initial velocity \( u = 0 \text{ m/s} \). For the truck: \( v_t = 0 + 2.10 \times 6.17 = 12.96 \text{ m/s} \). For the automobile: \( v_a = 0 + 3.40 \times 6.17 = 20.98 \text{ m/s} \).
05

Graph the positions as functions of time

Plot the position functions for both vehicles using the equations: truck: \( x_t(t) = \frac{1}{2} \times 2.10 \times t^2 \), automobile: \( x_a(t) = -24.88 + \frac{1}{2} \times 3.40 \times t^2 \). Both curves will intersect at \( t = 6.17 \text{ s} \) at \( x = 40.0 \text{ m} \).

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Key Concepts

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

Uniform Acceleration
Uniform acceleration occurs when the rate of acceleration remains constant over time. In this scenario, both the automobile and the truck experience uniform acceleration.

Here's the aspect of the uniform acceleration for each vehicle:
  • **The Truck:** With an acceleration of 2.10 m/s虏, the truck's velocity increases uniformly as time progresses. The displacement of the truck at any time can be calculated using the equation \[s = \frac{1}{2} a t^2,\]where \(s\) is the displacement, \(a\) is the acceleration, and \(t\) is the time.
  • **The Automobile:** Featuring a higher acceleration of 3.40 m/s虏, the automobile covers more ground faster compared to the truck. Its distance from the start is described by a similar equation \[s = -s_0 + \frac{1}{2} a t^2. \]Here, \(-s_0\) accounts for the initial positioning behind the truck.


Understanding how the uniform acceleration equations work allows us to solve for time or displacement using given variables. This is critical for predicting future positions or speeds of objects under consistent acceleration.
Relative Motion
Relative motion involves understanding how the position or velocity of one object compares to another. This concept is crucial when determining when and where two objects might meet. In this problem, we are analyzing when the automobile overtakes the truck.

  • The automobile and the truck start moving at the same instant, but since they have different accelerations, they will travel different distances over the same period.
  • The automobile begins a certain distance \(s_0\) behind the truck, which gives it an initial disadvantage.
  • To determine the time when the automobile overtakes the truck, we must find the moment both vehicles have traveled the same distance. This happens when their equations of motion yield the same displacement.
By assessing the relative motion, we can compute the initial distance and time of overlap, thus helping understand when and where two trajectories intersect.
Graphical Analysis
Graphical analysis helps in visualizing and interpreting the motion of the objects over time. By plotting the motion equations of both the automobile and the truck, we gain insight into their respective movements and meeting point.

  • **Position-Time Graph:** By graphing both the truck's and automobile's positions as time progresses, we can visually see their travel paths. Each line represents the motion equations derived earlier: \[x_t(t) = \frac{1}{2} \times 2.10 \times t^2\]for the truck, and \[x_a(t) = -24.88 + \frac{1}{2} \times 3.40 \times t^2\]for the automobile.
  • **Intersection Point:** The point where the two lines cross on the graph corresponds to the time and position at which the automobile overtakes the truck. This specific point should match our calculated time (around 6.17 seconds) and distance (40 meters).
Graphical representation not only aids in verifying calculations but also provides a clear overview of the entire motion dynamics involved in such kinematic problems.

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Most popular questions from this chapter

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