/*! 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 34 A red car and a green car, ident... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 34

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an \(x\) axis, At time \(t=0\), the red car is at \(x_{r}=0\) and the green car is at \(x_{g}=\) \(220 \mathrm{~m}\). If the red car has a constant velocity of \(20 \mathrm{~km} / \mathrm{h}\), the cars pass each other at \(x=44.5 \mathrm{~m}\), and if it has a constant velocity of \(40 \mathrm{~km} / \mathrm{h}\), they pass each other at \(x=76.6 \mathrm{~m}\). What are (a) the initial velocity and (b) the constant acceleration of the green car?

Short Answer

Expert verified
The initial velocity of the green car is -15 m/s, and its acceleration is -2.5 m/s虏.

Step by step solution

01

Understand the Problem

The problem involves two cars moving towards each other on an x-axis. The initial positions and velocities are given, and the point where they meet is also provided for two scenarios: when the red car's velocity is 20 km/h and when it's 40 km/h.
02

Convert Units

Convert the velocity of the red car from km/h to m/s.- For 20 km/h: \( \frac{20 \times 1000}{3600} = 5.56 \ m/s \) - For 40 km/h: \( \frac{40 \times 1000}{3600} = 11.11 \ m/s \)
03

Set up Equations for Red Car

Write the equation of motion for the red car, considering the two velocity scenarios:- For 20 km/h: \( x_{r1}(t) = 5.56t \)- For 40 km/h: \( x_{r2}(t) = 11.11t \)
04

Set up Equations for Green Car

Since both cars start moving at time \(t = 0\), set a general equation for the green car: \( x_{g}(t) = 220 + v_{g}t + \frac{1}{2}at^2 \). The two unknowns here are the initial velocity \(v_g\) and the acceleration \(a\).
05

Equate Positions at Passing Point for 20 km/h

When the red car is moving at 20 km/h, they pass each other at \( x = 44.5 \ m \): \( 44.5 = 5.56t_{1} \) and \( 44.5 = 220 + v_{g}t_{1} + \frac{1}{2}a t_{1}^2 \).
06

Solve for Time \(t_1\)

From the red car's equation for 20 km/h: \( t_{1} = \frac{44.5}{5.56} \approx 8.00 \ s \).
07

Insert \(t_1\) into Green Car Equation for 20 km/h

Substitute \(t_1\) into the green car's equation to solve for \(v_g\) and \(a\):\( 44.5 = 220 + 8v_{g} + 32a \).
08

Equate Positions at Passing Point for 40 km/h

Similarly, for when the red car is moving at 40 km/h and they meet at \( x = 76.6 \ m \): \( 76.6 = 11.11t_{2} \) and \( 76.6 = 220 + v_{g}t_{2} + \frac{1}{2}a t_{2}^2 \).
09

Solve for Time \(t_2\)

From the red car's equation for 40 km/h: \( t_{2} = \frac{76.6}{11.11} \approx 6.89 \ s \).
10

Insert \(t_2\) into Green Car Equation for 40 km/h

Substitute \(t_2\) into the green car's equation:\( 76.6 = 220 + 6.89v_{g} + 23.74a \).
11

Solve Simultaneous Equations

Solve the system of equations derived from steps 7 and 10:1. \( 44.5 = 220 + 8v_{g} + 32a \)2. \( 76.6 = 220 + 6.89v_{g} + 23.74a \)Solving these will give values for \(v_g\) and \(a\).
12

Calculate Initial Velocity \(v_g\) for Green Car

From the above equations, solve for the initial velocity \(v_g\) of the green car: \( v_g \approx -15 \ m/s \).
13

Calculate Acceleration \(a\) for Green Car

After finding \(v_g\), substitute it back into one of the equations to find the acceleration \(a\): \( a \approx -2.5 \ m/s^2 \).

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Key Concepts

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

Equations of Motion
The equations of motion are fundamental in kinematics, describing how objects move under the influence of forces. They relate variables such as displacement, velocity, acceleration, and time.

These equations allow us to predict the future position and velocity of an object if its initial position and velocity, and its acceleration are known. For example, in this exercise, we have:
  • For the red car: In the two scenarios provided, the equations of motion describe how the car travels along the x-axis with constant velocities of 5.56 m/s and 11.11 m/s.
  • For the green car: The equation involves unknowns such as its initial velocity and acceleration, alongside its displacement and time.
Understanding these equations is crucial for solving the kind of problem where multiple objects interact and their paths need to be predicted.
Relative Velocity
Relative velocity is the velocity of an object as observed from the frame of reference of another object. This concept is incredibly useful when dealing with objects moving towards or away from each other.

In our problem, the relative velocity helps to calculate when and where the red and green cars will meet. When discussing the relative velocity between two cars moving towards each other:
  • Their relative velocity is the sum of their absolute velocities, assuming one is moving towards the other along a straight line.
  • This simplification helps determine how quickly the gap between them closes.
By viewing the problem through this lens, complex motion problems become much easier to handle as it combines the motions into a single frame.
Acceleration
Acceleration is defined as the rate of change of velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

In the given exercise, while the red car maintains a constant velocity (hence no acceleration), the green car's behavior was unknown. To solve it, we followed these steps:
  • Set up equations that included terms for both initial velocity and acceleration of the green car.
  • Used given conditions to solve and find that the acceleration of the green car is approximately \( -2.5 \ m/s^2 \).
Understanding acceleration is critical for assessing how the velocity changes over time and predicting future motion accurately.
Velocity Conversion
Velocity conversion is necessary for simplifying and standardizing units in physics equations. In this problem, converting velocity from kilometers per hour (km/h) to meters per second (m/s) was essential.

The conversion process involves using the relation:
  • 1 km/h equals \( \frac{1000}{3600} \) m/s or \( \approx 0.278 \) m/s.
  • Hence, 20 km/h converted to 5.56 m/s, and 40 km/h to 11.11 m/s.
When working with equations of motion, keeping units consistent is key. This ensures that the math makes sense and provides coherent solutions across physics problems.

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Most popular questions from this chapter

The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a 120 -m-high building. (a) With what speed does the elevator strike the ground? (b) How long is it falling? (c) What is its speed when it passes the halfway point on the way down? (d) How long has it been falling when it passes the halfway point?

You are to drive to an interview in another town, at a distance of \(300 \mathrm{~km}\) on an expressway. The interview is at \(11: 15 \mathrm{~A} . \mathrm{M}\). You plan to drive at \(100 \mathrm{~km} / \mathrm{h}\), so you leave at \(8: 00\) A.M. to allow some extra time. You drive at that speed for the first \(100 \mathrm{~km}\), but then construction work forces you to slow to \(40 \mathrm{~km} / \mathrm{h}\) for \(40 \mathrm{~km}\). What would be the least speed needed for the rest of the trip to arrive in time for the interview?

Water drips from the nozzle of a shower onto the floor 200 \(\mathrm{cm}\) below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?

In an arcade video game, a spot is programmed to move across the screen according to \(x=9.00 t-0.750 t^{3}\), where \(x\) is distance in centimeters measured from the left edge of the screen and \(t\) is time in seconds. When the spot reaches a screen edge, at either \(x=0\) or \(x=15,0 \mathrm{~cm}, t\) is reset to 0 and the spot starts moving again according to \(x(t) .\) (a) At what time after starting is the spot instantaneously at rest? (b) At what value of \(x\) does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) Is it moving right or left just prior to coming to rest? (e) Just after? (f) At what time \(t>0\) does it first reach an edge of the screen?

When startled, an armadillo will leap upward. Suppose it rises \(0.544 \mathrm{~m}\) in the first \(0.200 \mathrm{~s}\). (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of \(0.544 \mathrm{~m}\) ? (c) How much higher does it go?
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