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Chapter 3: Problem 5

A pool ball is rolling along a table with a constant velocity. The components of its velocity vector are \(v_{x}=0.5 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=0.8 \mathrm{~m} / \mathrm{s} .\) Calculate the distance it travels in \(0.4 \mathrm{~s}\).

Short Answer

Expert verified
The ball travels approximately 0.377 meters.

Step by step solution

01

Identify the Velocity Components

The problem provides the horizontal velocity component as \(v_{x} = 0.5\text{ m/s}\) and the vertical velocity component as \(v_{y} = 0.8\text{ m/s}\). These are necessary for calculating the magnitude of the total velocity.
02

Calculate the Resultant Velocity

The magnitude of the resultant velocity \(v\) is calculated using the Pythagorean theorem: \[ v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(0.5)^2 + (0.8)^2} = \sqrt{0.25 + 0.64} = \sqrt{0.89} \approx 0.943\, \text{m/s} \]
03

Determine Distance Traveled

Use the formula for distance: \( \text{Distance} = \text{Velocity} \times \text{Time} \). The time is given as \(0.4\text{ s}\). Thus, the distance is: \[ \text{Distance} = 0.943 \text{ m/s} \times 0.4 \text{ s} = 0.3772 \text{ m} \]
04

Round the Final Answer

Round the calculated distance to three significant figures: \(0.377\text{ m}\).

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Key Concepts

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

Velocity Components
When we discuss the motion of an object, like a pool ball, it's crucial to understand its velocity components. Velocity is a vector quantity, which means it has both a magnitude and direction.
Components help us figure out the movement in each direction separately.

In this exercise, the ball's velocity is broken into two parts:
  • Horizontal component, denoted as \(v_{x}\), is 0.5 m/s.
  • Vertical component, denoted as \(v_{y}\), is 0.8 m/s.
These components form a right triangle, with the velocity components being the two sides.
This separation allows us to analyze complex movements more simply.
Resultant Velocity
After identifying the components of velocity, the next step is to calculate the ball's resultant velocity.
The resultant velocity represents the actual speed and direction of the ball along the table.

We use the Pythagorean Theorem to find the magnitude of this velocity:\[ v = \sqrt{v_{x}^2 + v_{y}^2} \]Substitute the given component values:\[ v = \sqrt{(0.5)^2 + (0.8)^2} \]Doing the math, we find:
  • The square of the horizontal component: \((0.5)^2 = 0.25\)
  • The square of the vertical component: \((0.8)^2 = 0.64\)
  • Add the squares: \(0.25 + 0.64 = 0.89\)
  • Finally, take the square root: \(\sqrt{0.89} \approx 0.943 \text{ m/s}\)
This resultant velocity shows us how the ball moves when both components act together.
Distance Calculation
The task is to find how far the pool ball travels in a given time.
We use the equation for distance, which is simply:\[ \text{Distance} = \text{Velocity} \times \text{Time} \]
In our example, the ball's resultant velocity is approximately 0.943 m/s, and time given is 0.4 s.
This leads us to:\[ \text{Distance} = 0.943 \text{ m/s} \times 0.4 \text{ s} = 0.3772 \text{ meters}\]By multiplying, we find the distance the ball travels over that period.
It's fascinating how simply we can use velocity and time to determine distance.
Pythagorean Theorem in Physics
The Pythagorean Theorem is a fundamental tool in physics, especially when dealing with vectors like velocity.
It explains the relationship between the sides of a right triangle.

In physics, it's often used to find the magnitude of a vector when we know its components.
This theorem states:\[ a^2 + b^2 = c^2 \]Where \(a\) and \(b\) are the triangle's shorter sides, and \(c\) is the hypotenuse.
  • Here, \(a\) and \(b\) represent the velocity components, \(v_{x}\) and \(v_{y}\).
  • \(c\) represents the resultant velocity.
By understanding this, we can easily break down complex vectors into simpler parts.
This makes calculations more manageable and helps predict how objects move in two dimensions.

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Most popular questions from this chapter

A tennis player hits a ball at ground level, giving it an initial velocity of \(24 \mathrm{~m} / \mathrm{s}\) at \(57^{\circ}\) above the horizontal. (a) What are the horizontal and vertical components of the ball's initial velocity? (b) How high above the ground does the ball go? (c) How long does it take the ball to reach its maximum height? (d) What are the ball's velocity and acceleration at its highest point? (e) For how long a time is the ball in the air? (f) When this ball lands on the court, how far is it from the place where it was hit?

A batter hits a baseball at a speed of \(35.0 \mathrm{~m} / \mathrm{s}\) and an angle of \(65.0^{\circ}\) above the horizontal. At the same instant, an outfielder 70.0 m away begins running away from the batter in the line of the ball's flight, hoping to catch it. How fast must the outfielder run to catch the ball? (Ignore air resistance, and assume the fielder catches the ball at the same height at which it left the bat.)

A water balloon slingshot launches its projectiles essentially from ground level at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). (a) At what angle should the slingshot be aimed to achieve its maximum range? (b) If shot at the angle you calculated in part (a), how far will a water balloon travel horizontally? (c) For how long will the balloon be in the air? (You can ignore air resistance.)

A man stands on the roof of a \(15.0-\mathrm{m}\) -tall building and throws a rock with a velocity of magnitude \(30.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. You can ignore air resistance. Calculate (a) the maximum height above the roof reached by the rock, (b) the magnitude of the velocity of the rock just before it strikes the ground, and (c) the horizontal distance from the base of the building to the point where the rock strikes the ground.

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of \(58.0^{\circ}\) above the horizontal, some of the tiny critters have reached a maximum height of \(58.7 \mathrm{~cm}\) above the level ground. (See Nature, Vol. 424,31 July \(2003,\) p. \(509 .\) ) (a) What was the takeoff speed for such a leap? (b) What horizontal distance did the froghopper cover for this world-record leap?
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