/*! 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 1 A car travels in the \(+x\) -dir... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(v_{\mathrm{av}-x}=6.25 \mathrm{m} / \mathrm{s} .\) How far does the car travel in 4.00 \(\mathrm{s}\) ?

Short Answer

Expert verified
The car travels 25.00 meters in 4.00 seconds.

Step by step solution

01

Identify Given Information

We are given the average velocity of the car, \( v_{\mathrm{av}-x} = 6.25 \text{ m/s} \), and the time interval, \( \Delta t = 4.00 \text{ s} \).
02

Understand the Formula for Distance

The formula for distance traveled when velocity and time are known is \( d = v_{\text{av}} \times \Delta t \), where \( d \) is the distance, \( v_{\text{av}} \) is the average velocity, and \( \Delta t \) is the time interval.
03

Substitute Values Into the Formula

Substitute the given values into the distance formula: \( d = 6.25 \text{ m/s} \times 4.00 \text{ s} \).
04

Calculate the Distance

Perform the multiplication to find the distance: \( d = 6.25 \times 4.00 = 25.00 \text{ meters} \).

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Key Concepts

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

Understanding Average Velocity
In kinematics, average velocity is an important concept that helps us understand how fast something is moving over a specific period of time. When we talk about average velocity, we're considering the overall rate of speed for an entire journey, instead of focusing on any particular instantaneous speed. In mathematical terms, average velocity can be calculated using the formula:
\[ v_{\text{av}} = \frac{\Delta x}{\Delta t} \]
where \( \Delta x \) represents the change in displacement and \( \Delta t \) is the time interval during which this change occurs. It's important to note that average velocity is a vector quantity, which means it has both magnitude (how much) and direction (which way). For example, moving in the positive x-direction at 6.25 m/s reflects both speed and direction. When given in a problem, this helps us calculate other quantities, like distance traveled, using simple mathematical operations.
Distance Calculation Simplified
Calculating distance when you know the average velocity and the time interval is straightforward and essential in kinematics. Here, the distance \( d \) can be found using the equation:
\[ d = v_{\text{av}} \times \Delta t \]
The logic behind this formula is straightforward: multiplying the average velocity by the time over which that velocity is maintained will give you the total distance covered. For instance, if a car travels at an average velocity of 6.25 m/s for 4 seconds, it will cover a distance of 25 meters. This calculation provides a clear picture of how far an object moves in a straight line when it maintains an average speed, without needing to account for fluctuations in speed during that period. Understanding this basic calculation is pivotal for solving various motion-related problems in physics.
Breaking Down the Time Interval
The concept of a time interval is crucial in kinematics as it denotes the duration over which a change in motion occurs. It is often represented by \( \Delta t \) and can be found by subtracting the initial time \( t_0 \) from the final time \( t_f \). So, the formula is:
\[ \Delta t = t_f - t_0 \]
Time intervals allow us to measure how long an object has been in motion, making it an essential component when calculating average velocity and distance. In the context of our problem, the time interval is straightforward, given as 4 seconds. This specific span tells us the car's duration of travel with a specified average speed. By clearly understanding time intervals, you can accurately measure and infer other kinetic quantities, as it links directly with both velocity and distance in formulating a complete description of motion.

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Most popular questions from this chapter

Entering the Freeway. A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 20 \(\mathrm{m} / \mathrm{s}(45 \mathrm{mi} / \mathrm{h})\) when it reaches the end of the \(120-\mathrm{m}\) -long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 \(\mathrm{m} / \mathrm{s} .\) What distance does the traffic travel while the car is moving the length of the ramp?

A subway train starts from rest at a station and accelerates at a rate of 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) for 14.0 \(\mathrm{s}\) . It runs at constant speed for 70.0 \(\mathrm{s}\) and slows down at a rate of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) until it stops at the next station. Find the total distance covered.

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?

In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 \(\mathrm{km}\) away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin in the nest and extend the \(+x\) -axis to the release point, what was the bird's average velocity in \(\mathrm{m} / \mathrm{s}(\) a) for the return flight, and (b) for the whole episode, from leaving the nest to returning?

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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