Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 s after it leaves the thrower's hand. Ignore air resistance. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs for the motion of the egg.

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.80 m to the bottom of the incline is 3.80 m/s. What is the speed of the block when it is 3.40 m from the top of the incline?

The fastest measured pitched baseball left the pitcher's hand at a speed of 45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

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?

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. (a) At what time after being ejected is the boulder moving at 20.0 m/s upward? (b) At what time is it moving at 20.0 m/s downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_y-t, v_y-t\), and \(y-t\) graphs for the motion.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free