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Chapter 2: Problem 83

Cars \(A\) and \(B\) travel in a straight line. The distance 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 the two cars 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
At the start, both cars are at the same point. They cross each other at time \(t\) where \(t\) is a solution of the equation \(\delta*t^{2} - (\gamma - \beta)*t + \alpha = 0\). The distance from car A to car B is neither increasing nor decreasing when the relative velocity of the two cars is zero, giving \(t = 0\) and \(t\) as solutions to \(3*\delta*t^{2} - (2*\gamma - 2*\beta)*t + \alpha = 0\). Both cars have the same acceleration when \(t = \frac{2*\gamma - 2*\beta}{6*\delta}\)

Step by step solution

01

Find out which car is ahead at the start

This can be done by substituting the initial time (i.e., \(t = 0\)) into both equations. For car A, \(x_{A}(0)=\alpha*0+\beta*0^{2} = 0 \ m\). For car B, we have \(x_{B}(0)=\gamma *0^{2}-\delta *0^{3} = 0 \ m\). Based on these computations, both cars are at the same point at the start.
02

Solve for the time(s) when the cars are at the same point

Set the two equations equal to each other and solve for \(t\). So, \( \alpha*t + \beta*t^{2} = \gamma*t^{2} - \delta*t^{3}\). This can be simplified to \delta*t^{3} - (\gamma - \beta)*t^{2} + \alpha*t = 0. By taking out the common factor, we get \(t * (\delta*t^{2} - (\gamma - \beta)*t + \alpha) = 0\). Therefore, the solution for \(t\) is either \(t = 0\) or solving for \(t\) from the equation \(\delta*t^{2} - (\gamma - \beta)*t + \alpha = 0\).
03

Calculate when the distance between cars A and B is neither increasing nor decreasing

This is when the rate of change of the distance between the two, i.e., their relative velocity, is zero. So, the first derivatives of the two functions have to be equal. By differentiating both equations with respect to \(t\), we get for car A: \(v_{A}(t) = \alpha + 2*\beta*t\) and for car B: \(v_{B}(t) = 2*\gamma*t - 3*\delta*t^{2}\). Setting \(v_{A}(t) = v_{B}(t)\) yields \( \alpha + 2*\beta*t = 2*\gamma*t - 3*\delta*t^{2}\). For the solution \(t\), we again set the equation \(t = 0\) and solve for \(t\) from the equation \(3*\delta*t^{2} - (2*\gamma - 2*\beta)*t + \alpha = 0\).
04

Determine when cars A and B have the same acceleration

Finally, compute when the two cars have the same acceleration by equating the second derivatives of both equations. Differentiating \( v_{A}(t) \) and \( v_{B}(t) \) with respect to \(t\), the acceleration of car A is \(a_{A}(t) = 2*\beta\) and of car B is \(a_{B}(t) = 2*\gamma - 6*\delta*t\). Equating \(a_{A}(t) = a_{B}(t)\) gives \(2*\beta = 2*\gamma - 6*\delta*t\). Solving for \(t\), we obtain \(t = \frac{2*\gamma - 2*\beta}{6*\delta}\).

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Key Concepts

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

Motion with Constant Acceleration
When studying kinematics, understanding motion with constant acceleration is fundamental. This type of motion refers to a scenario where an object's acceleration鈥攖he rate of change of its velocity鈥攔emains unchanged throughout the motion.

In a mathematical context, the acceleration, denoted as 'a', is a constant value. This simplicity allows us to use specific kinematics equations, sometimes called the 'equations of uniformly accelerated motion', which can predict future motion based on initial conditions. For example, if a car is accelerating at a steady rate, we can predict its future position and velocity after a certain time.

When solving problems involving constant acceleration, we use equations such as:
  • \( v = u + at \) (final velocity)
  • \( s = ut + \frac{1}{2}at^2 \) (displacement)
  • \( v^2 = u^2 + 2as \) (relation between velocity and displacement)
where:
  • \( v \) is final velocity,
  • \( u \) is initial velocity,
  • \( a \) is acceleration,
  • \( t \) is time,
  • \( s \) is displacement.
These equations are applicable to the motion of objects like cars moving down a straight road with steady speeding up or slowing down.
Relative Velocity
Relative velocity is a concept that often causes confusion but is crucial when examining the movement of two or more objects with respect to each other.

In essence, the relative velocity between two objects is the velocity of one object as perceived from the other. If you're on a train passing another train moving in the same direction, the relative velocity is the difference between your train's velocity and the other train's velocity.

To determine when two objects, such as cars A and B in the exercise, have a relative velocity of zero, we look for when the rate at which their distance from each other changes is zero. That is, their velocities are similar at a specific instant, which indicates they are moving together without getting closer or further apart.

Mathematically, we find the relative velocity by subtracting one's velocity from the others, and if interested in when this value equals zero, we would set their respective velocity equations equal to each other and solve for the time at which this condition is met.
Equations of Motion
Equations of motion are the bread and butter of solving kinematics problems. They connect the quantities of velocity, acceleration, time, and displacement in a motion that is either uniformly accelerated or decelerated.

A firm grasp of these equations is necessary to analyze and predict an object's behavior in motion. The most commonly used equations of motion are the ones previously mentioned, each serving a different purpose and providing a unique relationship between the kinematic quantities.

In cases where variable acceleration is at play, like that of car B in the exercise, these standard equations don't apply directly because we must consider the changing rates of acceleration. Instead, calculus methods such as integration and differentiation are used to find relationships between position, velocity, and acceleration over time.
Acceleration
Acceleration is a central concept in the study of motion and is defined as the rate of change of velocity. It can be thought of as how quickly an object speeds up, slows down, or changes direction.

Acceleration is a vector quantity, which means it has both a magnitude (how much) and a direction (which way). In a problem like the one with cars A and B, acceleration can be constant or it can change over time. With constant acceleration, as in the case of car A, the acceleration remains the same throughout the motion. However, for car B, acceleration changes as a function of time due to the term involving the third power of time.

To find the acceleration at any point in time, we can differentiate the velocity function with respect to time (since acceleration is the derivative of velocity). The equation for acceleration is generally denoted as \( a = \frac{dv}{dt} \) or, if the velocity is a function of position, as \( a = v \frac{dv}{dx} \). By understanding the concept of acceleration and its calculation, one can solve complex motion problems and determine how forces will affect the movement of objects.

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Most popular questions from this chapter

If the contraction of the left ventricle lasts \(250 \mathrm{~ms}\) and the speed of blood flow in the aorta (the large artery leaving the heart) is \(0.80 \mathrm{~m} / \mathrm{s}\) at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? (a) \(310 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{b}) 31 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{c}) 3.2 \mathrm{~m} / \mathrm{s}^{2} ;(\mathrm{d}) 0.32 \mathrm{~m} / \mathrm{s}^{2} .\)

An object's velocity is measured to be \(v_{x}(t)=\alpha-\beta t^{2}\), where \(\alpha=4.00 \mathrm{~m} / \mathrm{s}\) and \(\beta=2.00 \mathrm{~m} / \mathrm{s}^{3} .\) At \(t=0\) the object is at \(x=0 .\) (a) Calculate the object's position and acceleration as functions of time. (b) What is the object's maximum positive displacement from the origin?

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of \(3.50 \mathrm{~m} / \mathrm{s}^{2}\) upward. At \(25.0 \mathrm{~s}\) after launch, the second stage fires for \(10.0 \mathrm{~s}\), which boosts the rocket's velocity to \(132.5 \mathrm{~m} / \mathrm{s}\) upward at \(35.0 \mathrm{~s}\) after launch. This firing uses up all of the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Ignore air resistance. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stagetwo rocket be moving just as it reaches the launch pad?

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than \(1.00 \mathrm{~s}\) in the air (their "hang time"). Treat the athlete as a particle and let \(y_{\max }\) be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above \(y_{\max } / 2\) to the time it takes him to go from the floor to that height. Ignore air resistance.

A ball is thrown straight up from the ground with speed \(v_{0}\). At the same instant, a second ball is dropped from rest from a height \(H\) directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_{0}\) and \(g\) such that at the instant when the balls collide, the first ball is at the highest point of its motion.
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