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Chapter 2: Problem 6

A Honda Civic travels in a straight line along a road. The car's distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t) = \alpha{t^2} - \beta{t^3}\), where \(\alpha =\) 1.50 m/s\(^2\) and \(\beta =\) 0.0500 m/s\(^3\). Calculate the average velocity of the car for each time interval: (a) \(t =\) 0 to \(t =\) 2.00 s; (b) \(t =\) 0 to \(t =\) 4.00 s; (c) \(t =\) 2.00 s to \(t =\) 4.00 s.

Short Answer

Expert verified
(a) 2.8 m/s; (b) 5.2 m/s; (c) 7.6 m/s.

Step by step solution

01

Understand the Problem

The function given is \(x(t) = \alpha t^2 - \beta t^3\). We need to calculate the average velocity over the specified intervals. Average velocity, \( v_{avg} \), is defined as the total displacement divided by the total time taken, or \( v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}\).
02

Calculate x(t) for Each Time Interval (Part a)

For interval (a), \(t = 0\) to \(t = 2.00\) s, we calculate \(x(0)\) and \(x(2)\):- \(x(0) = \alpha (0)^2 - \beta (0)^3 = 0\)- \(x(2) = \alpha (2)^2 - \beta (2)^3 = 1.50 \times 4 - 0.0500 \times 8 = 6 - 0.4 = 5.6\) meters.
03

Calculate Average Velocity for Part a

Using the formula for average velocity, \(v_{avg} = \frac{x(2) - x(0)}{2 - 0}\):- \(v_{avg} = \frac{5.6 - 0}{2} = \frac{5.6}{2} = 2.8\) m/s.
04

Calculate x(t) for Each Time Interval (Part b)

For interval (b), \(t = 0\) to \(t = 4.00\) s, calculate \(x(0)\) and \(x(4)\):- \(x(0) = \alpha (0)^2 - \beta (0)^3 = 0\)- \(x(4) = \alpha (4)^2 - \beta (4)^3 = 1.50 \times 16 - 0.0500 \times 64 = 24 - 3.2 = 20.8\) meters.
05

Calculate Average Velocity for Part b

Using the formula for average velocity, \(v_{avg} = \frac{x(4) - x(0)}{4 - 0}\):- \(v_{avg} = \frac{20.8 - 0}{4} = \frac{20.8}{4} = 5.2\) m/s.
06

Calculate x(t) for Each Time Interval (Part c)

For interval (c), \(t = 2.00\) to \(t = 4.00\) s, calculate \(x(2)\) and \(x(4)\) (computed earlier):- \(x(2) = 5.6\) meters (already calculated in Part a)- \(x(4) = 20.8\) meters (already calculated in Part b).
07

Calculate Average Velocity for Part c

Using the formula for average velocity, \(v_{avg} = \frac{x(4) - x(2)}{4 - 2}\):- \(v_{avg} = \frac{20.8 - 5.6}{2} = \frac{15.2}{2} = 7.6\) m/s.

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It is a key concept in understanding how objects move from one place to another. When studying kinematics, we focus on several key variables:
  • Displacement - The change in position of an object.
  • Velocity - The rate of change of displacement with respect to time.
  • Acceleration - The rate of change of velocity over time.
Kinematics uses these concepts to predict the future position and velocity of an object. In this exercise, we analyze the Honda Civic's motion by examining how its position changes over time using the given displacement equation.
Distance-Time Relation
The distance-time relation is foundational in kinematics. It helps us understand how the distance traveled by an object is related to the time elapsed. For our exercise involving the Honda Civic, this relationship is represented by the equation:\[x(t) = \alpha t^2 - \beta t^3\]where the distance, or position, \( x(t) \) changes as time \( t \) progresses.
  • \( \alpha t^2 \) reflects the initial acceleration-driven change in position.
  • \( \beta t^3 \) depicts how the impact of acceleration diminishes over time due to the cubed time term.
By plugging in different values for \( t \), we can see how the position of the car changes at various times, forming a clear picture of the motion.
Honda Civic Motion
In this exercise, we're observing the motion of a Honda Civic as it travels in a straight line. The movement of the car is analyzed through the provided displacement equation. Let's break down the car's motion:- From rest, at \( t = 0 \), the car is at the origin.- As time passes (\( t = 2 \) seconds and \( t = 4 \) seconds), we calculate the car鈥檚 position using the equation and find how far it travels during these intervals.This simple yet powerful analysis allows us to understand its average velocity for any given time period. The average velocity tells us the uniform speed needed to cover the same distance in the same time frame.
Displacement Formula
The displacement formula used in our exercise is a mathematical representation of the motion of the Honda Civic. It's given by:\[x(t) = \alpha t^2 - \beta t^3\]In this formula:
  • \( \alpha \) represents the initial acceleration, measured in meters per second squared (m/s\(^2\)).
  • \( \beta \) denotes the rate at which acceleration impacts the motion over time, measured in meters per second cubed (m/s\(^3\)).
The displacement equation helps us determine the car's exact position at any time \( t \). By understanding displacement, we can calculate average velocity, which is crucial for determining how quickly the car is moving from one point to another. Average velocity is calculated as:\[v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}\]This gives us a practical measure of the car鈥檚 motion over a designated time period.

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Most popular questions from this chapter

A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s\(^2\). (a) For how much time and what distance does the student have to run at 5.0 m/s before she overtakes the bus? (b) When she reaches the bus, how fast is the bus traveling? (c) Sketch an \(x-t\) graph for both the student and the bus. Take \(x =\) 0 at the initial position of the student. (d) The equations you used in part (a) to find the time have a second solution, corresponding to a later time for which the student and bus are again at the same place if they continue their specified motions. Explain the significance of this second solution. How fast is the bus traveling at this point? (e) If the student's top speed is 3.5 m/s, will she catch the bus? (f) What is the \(minimum\) speed the student must have to just catch up with the bus? For what time and what distance does she have to run in that case?

A 7500-kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s\(^2\) and feels no appreciable air resistance. When it has reached a height of 525 m, its engines suddenly fail; the only force acting on it is now gravity. (a) What is the maximum height this rocket will reach above the launch pad? (b) How much time will elapse after engine failure before the rocket comes crashing down to the launch pad, and how fast will it be moving just before it crashes? (c) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs of the rocket's motion from the instant of blast-off to the instant just before it strikes the launch pad.

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.

A car travels in the \(+x\)-direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is \(v_{av-x}\) = 6.25 m/s. How far does the car travel in 4.00 s?

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_y =\) (2.80 m/s\(^3)t\), where the \(+y\)-direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t =\) 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?
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