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

Short Answer

Expert verified
The car travels 25 meters in 4 seconds.

Step by step solution

01

Understand the Problem

The car moves with an average velocity of 6.25 m/s in the +x-direction for 4.00 seconds. We need to calculate the distance traveled by the car within this time frame.
02

Recall the Distance Formula

The distance traveled by an object moving with constant velocity can be calculated using the formula:\[d = v_{av-x} \times t\]where \(d\) is the distance, \(v_{av-x}\) is the average velocity, and \(t\) is the time.
03

Substitute Known Values into the Formula

We have \(v_{av-x} = 6.25\, \text{m/s}\) and \(t = 4.00\, \text{s}\). Substitute these values into the formula:\[d = 6.25 \times 4.00\]
04

Perform the Calculation

Calculate the product:\[d = 25.00\, \text{meters}\]
05

Confirm the Units and Result

The units of our result are in meters, which is appropriate for measuring distance. The calculation confirms that the car travels a distance of 25 meters in 4 seconds.

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Key Concepts

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

Average Velocity
Average velocity is one of the key concepts in kinematics. It refers to the total displacement divided by the total time taken for that displacement. When we talk about displacement, we mean the change in position of an object. It is a vector quantity, which means it has both magnitude and direction. For example:
  • If an object moves from position A to position B, the average velocity tells us how fast and in which direction the object is moving overall.
  • In our exercise, the average velocity is given as 6.25 m/s in the +x-direction.
This means that over the 4.00 seconds that the car is traveling, it moves at an average speed of 6.25 meters every second along the positive x-axis. Average velocity is crucial for calculating distances when velocity changes over time because it simplifies these changes into a single measurement. However, if the velocity is constant, the average velocity is the same as the constant velocity of the object.
Distance Calculation
Calculating the distance traveled by an object is fundamental when it comes to understanding motion. The formula for distance when given average velocity and time is:\[d = v_{av-x} \times t\]This simple formula comes in handy when you know how fast an object is moving on average over a period of time. In this exercise:
  • The average velocity \(v_{av-x}\) is 6.25 m/s.
  • The time \(t\) is 4.00 seconds.
  • The formula becomes \(d = 6.25 \times 4.00\).
Following the calculation mentioned in the exercise, the car covers a distance of 25 meters in 4 seconds. It's important that each step of the calculation involves careful unit consideration to ensure the result is in the correct measurement, which is meters in this situation.
Constant Velocity
Constant velocity is a term that denotes motion at a fixed speed in a straight line. Unlike average velocity, constant velocity doesn't involve any changes over the time period considered. This can be described by a few points:
  • The speed does not change, so the object covers equal distances in equal intervals of time.
  • The direction of motion remains the same.
In the exercise, while it specifies average velocity, if that velocity were constant, it would mean the car maintains a steady speed of 6.25 m/s in the +x-direction. Constant velocity scenarios allow us to use straightforward calculations to determine things like distance traveled, using:\[d = vt\]where \(v\) is the constant velocity and \(t\) is the time. Constant velocity simplifies the calculation process, as you don’t have to account for any accelerations or decelerations.

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Most popular questions from this chapter

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

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of 960 m above the earth's surface. The rocket's engines give the rocket an upward acceleration of 16.0 m/s\(^2\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, air bags produce a maximum acceleration of \(60 g\) that lasts for only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g} ?\)

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.

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?

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