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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

In a physics lab experiment, you release a small steel ball at various heights above the ground and measure the ball's speed just before it strikes the ground. You plot your data on a graph that has the release height (in meters) on the vertical axis and the square of the final speed (in m\(^2\)/s\(^2\)) on the horizontal axis. In this graph your data points lie close to a straight line. (a) Using \(g\) = 9.80 m/s\(^2\) and ignoring the effect of air resistance, what is the numerical value of the slope of this straight line? (Include the correct units.) The presence of air resistance reduces the magnitude of the downward acceleration, and the effect of air resistance increases as the speed of the object increases. You repeat the experiment, but this time with a tennis ball as the object being dropped. Air resistance now has a noticeable effect on the data. (b) Is the final speed for a given release height higher than, lower than, or the same as when you ignored air resistance? (c) Is the graph of the release height versus the square of the final speed still a straight line? Sketch the qualitative shape of the graph when air resistance is present.

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?

A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by \(y(t) = b - ct + dt^2\) , where \(b =\) 800 m is the initial height of the lander above the surface, \(c =\) 60.0 m/s, and \(d =\) 1.05 m/s\(^2\). (a) What is the initial velocity of the lander, at \(t =\) 0? (b) What is the velocity of the lander just before it reaches the lunar surface?

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (\(\textbf{Fig. E2.44}\)). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does 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.

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?

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