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Chapter 2: Problem 63

Two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second. During each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. Thus, the velocity of each person changes from second to second. However, during any one second, the velocity remains constant. Make a position-time graph for one of the runners. From this graph, determine (a) how much time passes before the runners collide and (b) the speed with which each is running at the moment of collision.

Short Answer

Expert verified
Collision occurs at about 7.27 seconds; speed at collision is approximately 5.3 m/s.

Step by step solution

01

Define Initial Conditions

Let's first note that both runners start 100 meters apart. Both runners run 10 meters in the first second.
02

Determine the Distance Each Second

In the first second, each runner covers 10 meters. Every subsequent second, each covers 90% of the distance covered in the previous second. Therefore, in the second second, they each cover \(10 \times 0.9 = 9\) meters and in the third second, \(9 \times 0.9 = 8.1\) meters.
03

Calculate Cumulative Distance Covered by Both Runners

Add up the distances for each second to determine how far each runner has travelled. We find \(x(t)\) for every second \(t\) where \(x(t)\) is the cumulative distance covered.
04

Express Position as Function of Time

The position of the runner, starting at point 0, is given by \(x_1(t) = 10(1 + 0.9 + 0.9^2 + \ldots +0.9^{t-1})\). Similarly, the position of the runner from the opposite direction is \(x_2(t) = 100 - 10(1 + 0.9 + 0.9^2 + \ldots +0.9^{t-1})\).
05

Collision Condition

The runners collide when their positions are equal: \(x_1(t) = x_2(t)\). Specifically, when their cumulative distances add to 100 meters.
06

Use Geometric Series Formula

The series can be summed using the geometric series formula: \(S_n = a\frac{1-r^n}{1-r}\) where \(a = 10\) and \(r = 0.9\). For collision: \(2 \times 10 \frac{1-0.9^t}{1-0.9} = 100\).
07

Solve for Time of Collision

Simplify and solve \(20 \frac{1-0.9^t}{0.1} = 100\) to get \(1-0.9^t = 0.5\). Solving \(0.9^t = 0.5\) gives \(t \approx 7.27\) seconds.
08

Determine Speed at Collision

Since speed at any second is 90% of the last speed, calculate the speed at 7 seconds and slightly after (\(v_7 = 10 \times 0.9^6\)).

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Key Concepts

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

Geometric Series
In kinematics, when a quantity decreases by a constant ratio over equal time intervals, it forms a geometric series. This concept helps determine the distance each runner covers over time. Initially, each runner covers 10 meters in the first second.
Afterward, they cover 90% of the previous distance each second.
This sequence of distances, 10, 9, 8.1, and so on, forms a geometric series.
  • The first term, denoted as \(a\), is 10 meters.
  • The common ratio, \(r\), is 0.9.
To find the sum of the distances covered at any time \(t\), we use the formula for the sum of the first \(n\) terms of a geometric series: \[ S_n = a \frac{1-r^n}{1-r} \]This series allows us to calculate the cumulative distance covered until collision.
Position-Time Graph
A position-time graph visually represents how position changes with time. It plots time on the horizontal axis and position on the vertical axis. For the runner, the graph starts at 0 and shows a curve that represents the cumulative distance covered.
Each increment shows less distance as time progresses due to decreasing speed.
  • The runner's position during the first second is 10 meters.
  • In the second second, the position is 19 meters (10 meters + 9 meters).
By plotting these values, you create a decreasing exponential curve, illustrating the runner's position over time. The graph helps determine when the position of both runners becomes equal, indicating a collision.
Relative Motion
Relative motion is key in understanding two objects moving toward each other. In this scenario, both runners are moving with respect to one another. They start 100 meters apart and decrease this distance by running toward each other.
Understanding relative motion helps calculate when and where they collide. The relative velocity is the sum of both individual velocities since they move in opposite directions.
  • Initially, they collectively cover 20 meters per second (10 meters each).
  • Each subsequent second, this decreases by 10% of the previous total coverage.
By analyzing their relative motion, we solve when the sum of their covered distances equals the initial separation.
Velocity Calculation
Velocity is the speed of an object in a specific direction. In this problem, it is crucial to calculate the exact speed of each runner at any moment. During each second, velocity remains constant, changing only after a second passes.
The velocity at any given second is 90% of the velocity from the previous second.
  • For instance, after the first second, the velocity is 10 m/s.
  • After the second second, it is 9 m/s (10 m/s \( \times 0.9 \)).
To determine the speed at collision, calculate the velocity using the formula for the 7th second and a fraction after, \( v_7 = 10 \times 0.9^6 \). This demonstrates their constant deceleration up to the collision point.

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Most popular questions from this chapter

Consult Interactive Solution \(\underline{2} .17\) at before beginning this problem. A car is traveling along a straight road at a velocity of \(+36.0 \mathrm{~m} / \mathrm{s}\) when its engine cuts out. For the next twelve seconds the car slows down, and its average acceleration is \(\bar{a}_{1} .\) For the next six seconds the car slows down further, and its average acceleration is \(\bar{a}_{2}\). The velocity of the car at the end of the eighteen-second period is \(+28.0 \mathrm{~m} / \mathrm{s}\). The ratio of the average acceleration values is \(\bar{a}_{1} / \bar{a}_{2}=1.50 .\) Find the velocity of the car at the end of the initial twelve-second interval.

Consult Concept Simulation 2.1 at for help in preparing for this problem. A cheetah is hunting. Its prey runs for \(3.0 \mathrm{~s}\) at a constant velocity of \(+9.0 \mathrm{~m} / \mathrm{s}\). Starting from rest, what constant acceleration must the cheetah maintain in order to run the same distance as its prey runs in the same time?

Interactive Solution \(2.31\) at offers help in modeling this problem. A car is traveling at a constant speed of \(33 \mathrm{~m} / \mathrm{s}\) on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is \(2.5 \mathrm{~km}\) away?

For the first \(10.0 \mathrm{~km}\) of a marathon, a runner averages a velocity that has a magnitude of \(15.0 \mathrm{~km} / \mathrm{h}\). For the next \(15.0 \mathrm{~km}\), he averages \(10.0 \mathrm{~km} / \mathrm{h}\), and for the last \(15.0 \mathrm{~km},\) he averages \(5.0 \mathrm{~km} / \mathrm{h}\). Construct, to scale, the position-time graph for the runner.

A sprinter explodes out of the starting block with an acceleration of \(+2.3 \mathrm{~m} / \mathrm{s}^{2},\) which she sustains for \(1.2 \mathrm{~s}\). Then, her acceleration drops to zero for the rest of the race. What is her velocity (a) at \(t=1.2 \mathrm{~s}\) and \((\mathrm{b})\) at the end of the race?
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