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Chapter 2: Problem 7

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?

Short Answer

Expert verified
(a) 12 m/s, (b) 0 m/s at t=0, 15 m/s at t=5, 12 m/s at t=10, (c) 13.33 s.

Step by step solution

01

Define Average Velocity

The average velocity over a time interval is given by the formula \( v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} \). For this problem, we will use \( t_1 = 0 \) s and \( t_2 = 10 \) s.
02

Calculate Position at t=0 and t=10

At \( t = 0 \) s, \( x(0) = b(0)^2 - c(0)^3 = 0 \). At \( t = 10 \) s, substitute into the function: \( x(10) = b(10)^2 - c(10)^3 = 2.40 \times 100 - 0.120 \times 1000 \). So, \( x(10) = 240 - 120 = 120 \) m.
03

Calculate Average Velocity

Using \( x(0) = 0 \) m and \( x(10) = 120 \) m, the average velocity is \( v_{avg} = \frac{120 - 0}{10 - 0} = 12 \) m/s.
04

Define Instantaneous Velocity

The instantaneous velocity is the derivative of \( x(t) \) with respect to time, \( v(t) = \frac{dx(t)}{dt} = \frac{d}{dt}(b t^2 - c t^3) \). Calculating the derivative, \( v(t) = 2bt - 3ct^2 \).
05

Calculate Instantaneous Velocity at t=0, t=5, t=10

Substitute into the velocity equation: - At \( t=0 \) s: \( v(0) = 2b(0) - 3c(0)^2 = 0 \) m/s. - At \( t=5 \) s: \( v(5) = 2 \times 2.40 \times 5 - 3 \times 0.120 \times 25 = 24 - 9 = 15 \) m/s. - At \( t=10 \) s: \( v(10) = 2 \times 2.40 \times 10 - 3 \times 0.120 \times 100 = 48 - 36 = 12 \) m/s.
06

Determine When Car is at Rest Again

The car is at rest when this velocity equals zero: \( 2bt - 3ct^2 = 0 \).Factor out \( t \): \( t(2b - 3ct) = 0 \). This gives solutions of \( t = 0 \) and \( t = \frac{2b}{3c} \). Calculate the latter: \( t = \frac{2 \times 2.40}{3 \times 0.120} = \frac{4.8}{0.36} \approx 13.33 \) s.

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Key Concepts

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

average velocity
Average velocity helps us understand how fast an object is moving over a certain period of time. It is defined as the total change in position divided by the total time taken. The formula to find average velocity is given by:
  • \( v_{\text{avg}} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} \)
Here, \(x(t_2)\) and \(x(t_1)\) are the positions of the object at times \(t_2\) and \(t_1\) respectively. This calculation helps you to find out the overall performance of the car over the specified time.
This can be particularly useful when figuring out how different segments of a journey contribute to total travel time.
instantaneous velocity
Instantaneous velocity tells us about an object's speed and direction at a specific moment in time. Unlike average velocity, it is not calculated over an interval but at a single point. To find instantaneous velocity, we take the derivative of the position function with respect to time. The formula is:
  • \( v(t) = \frac{dx(t)}{dt} \)
By substituting specific values of time into this derivative, we can see exactly how fast the car is moving at that exact second.
This is helpful in scenarios where you need to know how speed changes in real-time, like in speed monitoring or rapid acceleration tests.
position function
The position function describes how an object's position changes over time. It is typically represented as \(x(t)\), where \(t\) is time. For our car problem, the position function is:
  • \(x(t) = b t^2 - c t^3\)
This function helps us predict where the car will be at any given time. Parameters \(b\) and \(c\) are constants that depend on the specifics of the car's movement.
Understanding the position function is essential for accurately planning travel, as it can show where an object will be at various points in time.
derivative of position
The derivative of the position function, often referred to as the velocity function, gives the rate of change of position with respect to time, which is the object's velocity. For the function \(x(t) = b t^2 - c t^3\), the derivative is:
  • \( v(t) = 2bt - 3ct^2 \)
This derivative provides the instantaneous velocity at any point \(t\). Calculating this helps in understanding how the speed of an object changes at any instant, providing deeper insight into its motion dynamics.
Derivatives are fundamental in calculus and physics, allowing us to better understand and predict how objects move in varying conditions.

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Most popular questions from this chapter

A flowerpot falls off a windowsill and falls past the window below. You may ignore air resistance. It takes the pot 0.420 s to pass from the top to the bottom of this window, which is 1.90 \(\mathrm{m}\) high. How far is the top of the window below the windowsill from which the flowerpot fell?

Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of \(H\) , how high (in terms of \(H )\) will the faster stone go? Assume free fall.

A world-class sprinter accelerates to his maximum speed in 4.0 s. He then maintains this speed for the remainder of a \(100-\mathrm{m}\) race, finishing with a total time of 9.1 \(\mathrm{s}\) . (a) What is the runner's average acceleration during the first 4.0 \(\mathrm{s} ?\) (b) What is his average acceleration during the last 5.1 \(\mathrm{s} ?\) (c) What is his average acceleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b).

Mars Landing. In January \(2004,\) NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages: Stage \(A:\) Friction with the atmosphere reduced the speed from \(19,300 \mathrm{km} / \mathrm{h}\) to 1600 \(\mathrm{km} / \mathrm{h}\) in 4.0 \(\mathrm{min.}\) Stage \(B:\) A parachute then opened to slow it down to 321 \(\mathrm{km} / \mathrm{h}\) in 94 \(\mathrm{s} .\) Stage \(C :\) Retro rockets then fired to reduce its speed to zero over a distance of 75 \(\mathrm{m}.\) Assume that each stage followed immediately after the preceding one and that the acceleration during each stage was constant. (a) Find the rocket's acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ) during each stage. (b) What total distance (in km) did the rocket travel during stages \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C} ?\)

On a 20 -mile bike ride, you ride the first 10 miles at an average speed of 8 \(\mathrm{mi} / \mathrm{h} .\) What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be (a) 4 \(\mathrm{mi} / \mathrm{h} ?\) (b) 12 \(\mathrm{mi} / \mathrm{h} ?\) (c) Given this average speed for the first 10 miles, can you possibly attain an average speed of 16 \(\mathrm{mi} / \mathrm{h}\) for the total 20 -mile ride? Explain.
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