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Chapter 2: Problem 36

Two soccer players start from rest, 48 \(\mathrm{m}\) apart. They run directly toward each other, both players accelerating. The first player's acceleration has a magnitude of 0.50 \(\mathrm{m} / \mathrm{s}^{2}\) . The second player's acceleration has a magnitude of 0.30 \(\mathrm{m} / \mathrm{s}^{2}\) (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

Short Answer

Expert verified
The players collide after approximately 10.95 seconds; Player 1 runs 30 meters.

Step by step solution

01

Stepively Understand the Problem

We have two players running toward each other, starting from rest and accelerating at given rates. We need to find when and where they will meet. Player 1 has an acceleration of 0.50 m/s虏, and Player 2 has an acceleration of 0.30 m/s虏. The initial distance between them is 48 m.
02

Set Up Equations of Motion

Both players start from rest, so their initial velocities are 0. Using the equation of motion for position from rest under constant acceleration, we have:For Player 1: \[x_1 = \frac{1}{2} a_1 t^2\]For Player 2:\[x_2 = \frac{1}{2} a_2 t^2\]where \( a_1 = 0.50 \, \mathrm{m/s}^2 \) and \( a_2 = 0.30 \, \mathrm{m/s}^2 \).
03

Express the Meeting Condition

Since they run toward each other, the sum of the distances they cover will equal the initial separation distance, 48 m:\[x_1 + x_2 = 48\]Substitute the expressions for \( x_1 \) and \( x_2 \):\[\frac{1}{2} a_1 t^2 + \frac{1}{2} a_2 t^2 = 48\]
04

Solve for Time

Combine the terms:\[\frac{1}{2} (a_1 + a_2) t^2 = 48\]Insert the values for \( a_1 \) and \( a_2 \):\[\frac{1}{2} (0.50 + 0.30) t^2 = 48\]\[0.40 t^2 = 48\]Solve for \( t^2 \):\[t^2 = \frac{48}{0.40}\]\[t^2 = 120\]\[t = \sqrt{120} \approx 10.95 \, \text{s}\]
05

Calculate the Distance Run by Player 1

Use the first player's position equation:\[x_1 = \frac{1}{2} a_1 t^2\]Substitute the known values:\[x_1 = \frac{1}{2} (0.50) (10.95)^2\]Calculate:\[x_1 = 0.25 \times 120\]\[x_1 = 30 \, \text{m}\]
06

Step Meet

Players collide in approximately 10.95 seconds after starting, and by then, the first player has run 30 meters.

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Key Concepts

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

Acceleration
Acceleration is a key concept in kinematics, describing how quickly an object changes its velocity. It's expressed in meters per second squared (\(\text{m/s}^2\)). In this scenario, both soccer players have different accelerations. Player 1 has an acceleration of 0.50 \(\text{m/s}^2\), while Player 2 has an acceleration of 0.30 \(\text{m/s}^2\). This means Player 1 increases their speed faster than Player 2. Acceleration is crucial in determining how soon two objects moving toward each other will meet because it directly affects how fast they cover the initial distance between them.
Motion
Motion refers to the change in position of an object over time. In this exercise, both players start from rest, meaning their initial velocities are zero. As they accelerate towards each other, they experience a change in motion. The initial distance of 48 meters between them decreases over time as their speeds increase. Understanding their motion is essential to calculating when and where they will collide.
Equations of Motion
The equations of motion are a set of formulas used to calculate different parameters like displacement, velocity, and acceleration when an object is in motion. The primary equation used here is:\[x = \frac{1}{2} a t^2\]Here, \(x\) is the displacement, \(a\) is the acceleration, and \(t\) is the time. Both players' motions are described using this equation as they start from rest. By setting up equations for each player's motion, we can establish when the sum of their displacements equals the initial distance of 48 meters, leading us to solve for the collision time.
Relative Velocity
Relative velocity describes how one object moves concerning another. Since the players run towards each other, their relative velocity simply adds up due to their respective accelerations. While each player's velocity is increasing individually, their combined speed in relation to each other grows even faster. Situations like this, where two objects move toward each other, are classic examples of relative velocity in action. Understanding how to compute this allows us to predict their meeting time efficiently.

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Most popular questions from this chapter

In 1954 the English runner Roger Bannister broke the four-minute barrier for the mile with a time of \(3 : 59.4 \mathrm{s}(3 \mathrm{min} \text { and } 59.4 \mathrm{s}) .\) In 1999 the Moroccan runner Hicham el-Guerrouj set a record of \(3 : 43.13\) s for the mile. If these two runners had run in the same race, each running the entire race at the average speed that earned him a place in the record books, el-Guerrouj would have won. By how many meters?

The leader of a bicycle race is traveling with a constant velocity of \(+11.10 \mathrm{m} / \mathrm{s}\) and is 10.0 \(\mathrm{m}\) ahead of the second-place cyclist. The second-place cyclist has a velocity of \(+9.50 \mathrm{m} / \mathrm{s}\) and an acceleration of \(+1.20 \mathrm{m} / \mathrm{s}^{2}\) . How much time elapses before he catches the leader?

A car is traveling at \(20.0 \mathrm{m} / \mathrm{s},\) and the driver sees a traffic light turn red. After 0.530 s (the reaction time), the driver applies the brakes, and the car decelerates at 7.00 \(\mathrm{m} / \mathrm{s}^{2} .\) What is the stopping distance of the car, as measured from the point where the driver first sees the red light?

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{m} / \mathrm{s},\) and during a 5.0 \(\mathrm{s}\) - interval, it changes to a final speed of (a) 29.0 \(\mathrm{m} / \mathrm{s}\) and \((\mathrm{b}) 23.0 \mathrm{m} / \mathrm{s}\) . In each case, find the acceleration ( magnitude and algebraic sign) and state whether or not the car is decelerating.

A jogger accelerates from rest to 3.0 \(\mathrm{m} / \mathrm{s}\) in 2.0 s. A car accelerates from 38.0 to 41.0 \(\mathrm{m} / \mathrm{s}\) also in 2.0 \(\mathrm{s}\) . (a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the 2.0 \(\mathrm{s} ?\) If so, how much farther?
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