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Chapter 2: Problem 95

Two cars, \(A\) and \(B,\) travel in a straight line. The dis tance of \(A\) from the starting point is given as a function of time by \(x_{A}(t)=\alpha t+\beta t^{2},\) with \(\alpha=2.60 \mathrm{m} / \mathrm{s}\) and \(\beta=1.20 \mathrm{m} / \mathrm{s}^{2} .\) The distance of \(B\) from the starting point is \(x_{B}(t)=\gamma t^{2}-\delta t^{3},\) with \(\gamma=2.80 \mathrm{m} / \mathrm{s}^{2}\) and \(\delta=0.20 \mathrm{m} / \mathrm{s}^{3} .\) (a) Which car is ahead just after they leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from \(A\) to \(B\) neither increasing nor decreasing? (d) At what time(s) do \(A\) and \(B\) have the same acceleration?

Short Answer

Expert verified
(a) Neither, (b) 0 s and 2 s, (c) 0.65 s and 6.65 s, (d) 2.67 s.

Step by step solution

01

Initial Distance Analysis

First, calculate the initial distance for both cars when they leave the starting point (i.e., when \( t = 0 \)). For car \( A \), the distance function is given by \( x_A(t) = \alpha t + \beta t^2 \). For \( t = 0 \), \( x_A(0) = 0 \). For car \( B \), the distance function is \( x_B(t) = \gamma t^2 - \delta t^3 \). At \( t = 0 \), \( x_B(0) = 0 \). Both cars are at the starting point when \( t = 0 \). Thus, neither car is ahead just after they leave the starting point.
02

Finding Intersection Points

To find the times at which both cars are at the same point, we equate their distance functions: \( x_A(t) = x_B(t) \). This gives the equation \[ \alpha t + \beta t^2 = \gamma t^2 - \delta t^3 \]. Substituting the values, the equation becomes \[ 2.60t + 1.20t^2 = 2.80t^2 - 0.20t^3 \]. Simplifying, \[ 0.20t^2 - 0.20t^3 - 2.60t = 0 \], we factor out \( t \): \[ t(0.20t^2 + 0.20t - 2.60) = 0 \]. Thus, \( t = 0 \) or solve \( 0.20t^2 + 0.20t - 2.60 = 0 \), using the quadratic formula: \[ t = \frac{-0.20 \pm \sqrt{(0.20)^2 - 4 \times 0.20 \times (-2.60)}}{2 \times 0.20} \], solving gives \( t = 2 \text{ s and } -6.5 \text{ s} \). Since time cannot be negative, \( t = 0 \) and \( t = 2 \text{ s} \) are the times at which the cars intersect.
03

Analyzing Relative Motion

The relative velocity \( v_{A,B}(t) \) of car \( A \) to \( B \) is given by \( v_A(t) - v_B(t) \), where \( v_A(t) = \frac{d}{dt} x_A(t) = 2.60 + 2.4t \) and \( v_B(t) = \frac{d}{dt} x_B(t) = 5.60t - 0.60t^2 \). Hence, \( v_{A,B}(t) = 2.60 + 1.20t - (5.60t - 0.60t^2) = 2.60 - 4.40t + 0.60t^2 \). Setting \( v_{A,B}(t) = 0 \) to find when the distance is neither increasing nor decreasing: \( 0.60t^2 - 4.40t + 2.60 = 0 \). Using the quadratic formula yields \( t = \frac{4.40 \pm \sqrt{4.40^2 - 4 \times 0.60 \times 2.60}}{2 \times 0.60} \), resulting in \( t \approx 0.65 \text{ s and } 6.65 \text{ s} \). These are the times when the distance from \( A \) to \( B \) is static.
04

Identifying Acceleration Equality

Determine when the accelerations are equal. The acceleration function for car \( A \) is \( a_A(t) = \frac{d^2}{dt^2}x_A(t) = 2.40 \). For car \( B \), it is \( a_B(t) = \frac{d^2}{dt^2}x_B(t) = 5.60 - 1.20t \). Set \( a_A(t) = a_B(t) \), yielding \( 2.40 = 5.60 - 1.20t \). Solving this equation gives \[ 1.20t = 5.60 - 2.40 \], resulting in \( 1.20t = 3.20 \), thus \( t = \frac{3.20}{1.20} = 2.67 \text{ s} \). This is the time when both cars have the same acceleration.

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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 crucial for analyzing the paths of moving objects, like cars in this exercise. They help determine position, velocity, and acceleration at any given time.

For car A, the position function is given as \( x_A(t) = \alpha t + \beta t^2 \), where \( \alpha = 2.60 \text{ m/s} \) and \( \beta = 1.20 \text{ m/s}^2 \). This formula combines constant velocity \( \alpha \) and constant acceleration \( \beta \) terms.

For car B, the position function is \( x_B(t) = \gamma t^2 - \delta t^3 \), with \( \gamma = 2.80 \text{ m/s}^2 \) and \( \delta = 0.20 \text{ m/s}^3 \). Here, the distance involves both quadratic and cubic terms, indicating changing acceleration.

By equating \( x_A(t) \) with \( x_B(t) \), we can solve for the times when both cars occupy the same position. This provides valuable insights into how the motion of each vehicle compares over time.
Relative Velocity
Relative velocity is the velocity of one object as seen from another moving object. It's like looking at how fast one car is moving relative to another car.

To find the relative velocity of car A to car B, calculate \( v_{A,B}(t) = v_A(t) - v_B(t) \). Here, \( v_A(t) = 2.60 + 2.4t \) is derived from differentiating \( x_A(t) \), and \( v_B(t) = 5.60t - 0.60t^2 \) is from differentiating \( x_B(t) \).

This equation simplifies to \( v_{A,B}(t) = 2.60 - 4.40t + 0.60t^2 \). Setting \( v_{A,B}(t) = 0 \) can find the times when the cars are neither moving closer nor farther apart. In this exercise, it tells us when the distance between the cars is constant over time.

Understanding relative velocity is critical for analyzing interactions between objects moving along the same path or in a similar direction.
Acceleration
Acceleration is the rate of change of velocity over time. It's a vector, meaning it has both a magnitude and direction.

For car A, the acceleration is a constant \( a_A(t) = 2.40 \text{ m/s}^2 \) since the equation \( x_A(t) \) is quadratic, indicating uniform acceleration.

For car B, acceleration changes with time and is given by \( a_B(t) = 5.60 - 1.20t \). This function shows that car B's acceleration decreases linearly over time.

Equalizing the two accelerations, \( a_A(t) = a_B(t) \), identifies when the cars share the same acceleration in this exercise. With uniform and changing acceleration for cars A and B, respectively, this point of equality occurs at a specific time, helping us understand how their movement patterns converge.

Grasping the concept of acceleration aids in predicting future positions and velocities based on the current state of motion.

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Most popular questions from this chapter

CALC A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by \(x(t)=b t^{2}-c t^{3},\) where \(b=2.40 \mathrm{m} / \mathrm{s}^{2}\) and \(c=0.120 \mathrm{m} / \mathrm{s}^{3} .\) (a) Calculate the average velocity of the car for the time interval \(t=0\) to \(t=10.0\) s. (b) Calculate the instantaneous velocity of the car at \(t=0, t=5.0 \mathrm{s},\) and \(t=10.0 \mathrm{s}\) . (c) How long after starting from rest is the car again at rest?

The Fastest (and Most Expensive) Car! The table shows test data for the Bugatti Veyron, the fastest car made. The car is moving in a straight line (the \(x\)-axis). \(\begin{array}{llll}{\text { Time }(s)} & {0} & {2.1} & {20.0} & {53} \\\ {\text { Speed (mijh) }} & {0} & {60} & {200} & {253}\end{array}\) (a) Make a \(v_{x}-t\) graph of this car's velocity \((\) in \(\mathrm{mi} / \mathrm{h})\) as a function of time. Is its acceleration constant? (b) Calculate the car's average acceleration ( in \(\mathrm{m} / \mathrm{s}^{2} )\) between ( i 0 and \(2.1 \mathrm{s} ;\) (ii) 2.1 \(\mathrm{s}\) and 20.0 \(\mathrm{s}\); (iii) 20.0 s and 53 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300 will be built, it runs out of gas in 12 minutes at top speed, and it costs \(\$ 1.25\) million!)

Launch of the Space Shuttle. At launch the space shuttle weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 \(\mathrm{km} / \mathrm{h}\) , and at the end of the first 1.00 \(\mathrm{min}\) its speed is 1610 \(\mathrm{km} / \mathrm{h}\) . (a) What is the average acceleration (\(\operatorname{in}\) \(\mathrm{m} / \mathrm{s}^{2}\) ) of the shuttle (i) during the first \(8.00 \mathrm{s},\) and (ii) between 8.00 \(\mathrm{s}\) and the end of the first 1.00 \(\mathrm{min}\) ? (b) Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals, what distance does the shuttle travel (i) during the first \(8.00 \mathrm{s},\) and ( ii) during the interval from 8.00 s to 1.00 \(\mathrm{min}\)?

Starting from the front door of your ranch house, you walk 60.0 \(\mathrm{m}\) due east to your windmill, and then you turn around and slowly walk 40.0 \(\mathrm{m}\) west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from your front door to the bench, what are (a) your average velocity and (b) your average speed?

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 \(\mathrm{m} / \mathrm{s}\) releases a sandbag at an instant when the balloon is 40.0 \(\mathrm{m}\) above the ground (Fig. E2.44). After it is released, the sandbag is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 after its release. (b) How many seconds after its release will the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.
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