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Chapter 2: Problem 5

Two runners start simultaneously from the same point on a circular \(200-m\) track and \(\mathrm{run}\) in opposite directions. One runs at a constant speed of 6.20 \(\mathrm{m} / \mathrm{s}\) , and the other runs at a constant speed of 5.50 \(\mathrm{m} / \mathrm{s}\) . When they first mect, (a) for how long a time will they have been running, and (b) how far will each one have run along the track?

Short Answer

Expert verified
They meet after approximately 17.09 seconds. The first runner runs about 105.96 meters, and the second runner runs about 95.00 meters.

Step by step solution

01

Understanding the Problem

The runners are moving in opposite directions on a 200m track. Each is running at a constant speed. We need to find out when they first meet and the distance each ran by then.
02

Calculating Relative Speed

Since the runners are moving in opposite directions, their relative speed is the sum of their individual speeds: 6.20 m/s + 5.50 m/s = 11.70 m/s.
03

Determining Meeting Time

To find the time when they first meet, consider that they will meet when their combined distance equals the track's circumference. The equation is: \(200 = 11.70 \times t\) where \(t\) is the time in seconds. Solving for \(t\), we get \(t = \frac{200}{11.70}\).
04

Solving for Time

Calculate \(t = \frac{200}{11.70} \approx 17.09\) seconds. Thus, they meet after approximately 17.09 seconds.
05

Calculating Distance Run by Each Runner

For the first runner: \( \text{distance} = 6.20 \times 17.09 \approx 105.96 \) meters. For the second runner: \(\text{distance} = 5.50 \times 17.09 \approx 94.995\) meters.

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Key Concepts

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

Circular Motion
Circular motion refers to the movement of an object along a circular path. In the case of the runners, they are moving around a circular track. This type of motion is unique because the direction of the object's velocity is constantly changing, even though if the speed鈥攖he magnitude of the velocity鈥攔emains constant.
In our track example, both runners start at the same point but head in opposite directions around a 200-meter circular path. As they move, they each experience circular motion along their respective routes. Here are a few things to remember about circular motion:
  • It's important to note that the path's curvature keeps changing the direction of motion.
  • Consequently, even if speed is constant, there is continuous acceleration due to the change in direction.
  • In problems involving circular tracks, imagining the motion as occurring along the perimeter of a circle can help visualize the scenario better.
Distance and Displacement
Distance and displacement are fundamental concepts in understanding motion. Distance refers to how much ground an object has covered, regardless of its starting or ending point. Displacement measures the object's overall change in position, taking direction into account. In the context of circular motion and the runners:
  • Distance is the total length of the path each runner takes until they meet. It's simply calculated by speed multiplied by time.
  • Displacement would technically be zero upon their first meeting point, as they both started and ended their journey at the same spot in terms of direction (they both returned to the start after one lap in opposite directions).
Thus, in circular tracks, while the distance continues to increase as runners move, their displacement can often remain zero or depend on their current position relative to the starting point. This nuance is vital when solving problems in physics involving motion.
Constant Speed
When an object moves with constant speed, it covers equal distances in equal intervals of time without accelerating or decelerating. In the current scenario, each runner maintains a consistent pace鈥�6.20 m/s and 5.50 m/s, respectively. Constant speed does not imply constant velocity. Since velocity includes both speed and direction, in circular motion, the velocity keeps changing due to the directional change. Here's why constant speed is important:
  • If a runner maintains a constant speed, it becomes easier to make predictions about their position and meeting times.
  • Calculations become straightforward, as the formula for distance traveled at constant speed is simply the product of speed and time.
This simplifying assumption allows us to predict motion outcomes, like determining the point of their first meeting on the track, using straightforward calculations.

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Most popular questions from this chapter

A marble is released from one rim of a hemispherical bowl of diameter 50.0 \(\mathrm{cm}\) and rolls down and up to the opposite rim in 10.0 s. Find (a) the uverage speed and (b) the average velocity of the marble.

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) upward. At 25.0 \(\mathrm{s}\) after launch, the rocket fires the second stage, which suddenly boosts its speed to 132.5 \(\mathrm{m} / \mathrm{s}\) upward. This firing uses up all the fuel, however, so then the only force acting on the rocket is gravity. Air resistance is negligible. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stage-two rocket be moving just as it reaches the launch pad?

A ball is thrown straight up from the ground with speed \(v_{0}\) . At the same instant, a second ball is dropped from rest from a height \(H\) , directly above the point where the first ball was thrown upward. There is no air resistance, (a) Find the time at which the two balls collide. (b) Find the value of \(H\) in terms of \(v_{0}\) and \(g\) so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Sam heaves a \(16-\) - Ib shot straight upward, giving it a constant upward accleration from rest of 45.0 \(\mathrm{m} / \mathrm{s}^{2}\) for 64.0 \(\mathrm{cm} .\) He releases it 2.20 \(\mathrm{m}\) above the ground. You may ignore air resistance. \((\mathrm{a})\) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 \(\mathrm{m}\) above the ground?

A hot-air balloonist, rising vertically with a constant velocity of magnitude \(5.00 \mathrm{m} / \mathrm{s},\) releases a sandbag at an instant when the balloon is 40.0 \(\mathrm{m}\) above the ground (Fig. 2.41\()\) . After it is released, the sandbag is in frec 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 s strike after its release will 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.
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