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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

An object is moving along the \(x\)-axis. At \(t =\) 0 it has velocity \(v_{0x}\) = 20.0 m/s. Starting at time \(t =\) 0 it has acceleration \(a_x = -Ct\), where \(C\) has units of m/s\(^3\). (a) What is the value of \(C\) if the object stops in 8.00 s after \(t =\) 0? (b) For the value of \(C\) calculated in part (a), how far does the object travel during the 8.00 s?

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

In the fastest measured tennis serve, the ball left the racquet at 73.14 m/s. A served tennis ball is typically in contact with the racquet for 30.0 ms and starts from rest. Assume constant acceleration. (a) What was the ball's acceleration during this serve? (b) How far did the ball travel during the serve?

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What are (a) its speed at the first point and (b) its acceleration?

An object's velocity is measured to be \(v_x(t) =\alpha - \beta{t}^2\), where \(\alpha\) = 4.00 m/s and \(\beta\) = 2.00 m/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?

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