/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A speed ramp at an airport is a ... [FREE SOLUTION] | 91Ó°ÊÓ

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A speed ramp at an airport is a moving conveyor belt on which you can either stand or walk. It is intended to get you from place to place more quickly. Suppose a speed ramp is \(120 \mathrm{~m}\) long. When you walk at a comfortable speed on the ground, you cover this distance in \(86 \mathrm{~s}\). When you walk on the speed ramp at this same comfortable speed, you cover this distance in 35 s. Determine the speed at which the speed ramp is moving relative to the ground.

Short Answer

Expert verified
The speed of the speed ramp relative to the ground is approximately 2.034 m/s.

Step by step solution

01

Determine Walking Speed on Ground

First, calculate the walking speed on the ground using the formula for speed: speed equals distance divided by time. Given:- Distance covered on ground = 120 m- Time taken on ground = 86 sWalking speed on the ground = \(\frac{120 \text{ m}}{86 \text{ s}}\) = 1.395 m/s (approximately)
02

Determine Walking Speed on Speed Ramp

Next, calculate the speed when walking on the speed ramp using the same formula.Given:- Distance covered on the speed ramp = 120 m- Time taken on the speed ramp = 35 sWalking speed on the speed ramp = \(\frac{120 \text{ m}}{35 \text{ s}}\) = 3.429 m/s (approximately)
03

Calculate Speed of the Speed Ramp

Now, find the speed of the speed ramp relative to the ground. It is the difference between the walking speed on the speed ramp and the walking speed on the ground. Speed of the speed ramp = walking speed on the speed ramp - walking speed on the ground = 3.429 m/s - 1.395 m/s = 2.034 m/s

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Key Concepts

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

Speed Calculation
Calculating speed is a straightforward process that involves determining how fast an object is moving. Speed is measured as the distance traveled divided by the time it takes to travel that distance. The general formula used for speed calculation is:\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]This simple formula allows us to find the speed of someone walking, driving, or even a conveyor belt at an airport! By knowing the distance covered and the time taken, we can easily find out how fast an object is moving. For example, if you walk on solid ground and cover a distance of 120 meters in 86 seconds, your walking speed is:\[ \text{Walking Speed} = \frac{120 \text{ m}}{86 \text{ s}} \approx 1.395 \text{ m/s} \] Here, "m/s" stands for meters per second, which is a common unit of speed in sciences.
Distance and Time Relationship
Understanding the relationship between distance and time is key to mastering concepts like speed and velocity. When we calculate speed, we directly link how far something travels (distance) with how long it takes to get there (time). This relationship can be expressed in the equation we used earlier:\[ \text{Distance} = \text{Speed} \times \text{Time} \]This means that if you know two of the three variables (distance, speed, time), you can find the third.
  • If you double your speed (and all other factors remain the same), you need half the time to cover the same distance.
  • If the distance is increased while time remains constant, the speed must be higher to cover that extra distance.
In our example, using the speed ramp reduces the time you need to reach the other end, showing how the speed of the ramp contributes to decreasing the time needed for the same distance.
Velocity and Speed Concepts
While speed and velocity are often used interchangeably in casual conversation, they have specific meanings in physics. Speed is a scalar quantity meaning it only describes how fast an object is moving regardless of direction. It is purely based on magnitude. On the other hand, velocity is a vector quantity, which means it involves both a magnitude and a direction. Knowing both gives you a complete picture of motion. For example:
  • If you walk 3 m/s in a straight line north, your velocity includes the speed (3 m/s) and direction (north).
  • However, if you're just asked for speed, you'd mention the 3 m/s, omitting the direction.
In our airport ramp scenario, although we've primarily focused on speed, any slight scrutiny involving direction changes would include velocity for precise navigation understanding.

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Most popular questions from this chapter

A small aircraft is headed due south with a speed of \(57.8 \mathrm{~m} / \mathrm{s}\) with respect to still air. Then, for \(9.00 \times 10^{2}\) s a wind blows the plane so that it moves in a direction \(45.0^{\circ}\) west of south, even though the plane continues to point due south. The plane travels \(81.0 \mathrm{~km}\) with respect to the ground in this time. Determine the velocity (magnitude and direction) of the wind with respect to the ground. Determine the directional angle relative to due south.

(a) A projectile is launched at a speed \(v_{0}\) and at an angle above the horizontal; its initial velocity components are \(v_{0_{x}}\) and \(v_{0_{y}}\). In the absence of air resistance, what is the speed of the projectile at the peak of its trajectory? (b) What is its speed just before it lands at the same vertical level from which it was launched? (c) Consider its speed at a point that is at a vertical level between that in (a) and (b). How does the speed at this point compare with the speeds identified in (a) and (b)? In each case, give your reasoning. A golfer hits a shot to a green that is elevated \(3.0 \mathrm{~m}\) above the point where the ball is struck. The ball leaves the club at a speed of \(14.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(40.0^{\circ}\) above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands. Check to see that your answer is consistent with your answer to part (c) of the Concept Questions.

A fire hose ejects a stream of water at an angle of \(35.0^{\circ}\) above the horizontal. The water leaves the nozzle with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

In the javelin throw at a track-and-field event, the javelin is launched at a speed of \(29 \mathrm{~m} /\) s at an angle of \(36^{\circ}\) above the horizontal. As the javelin travels upward, its velocity points above the horizontal at an angle that decreases as time passes. How much time is required for the angle to be reduced from \(36^{\circ}\) at launch to \(18^{\circ}\) ?

A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of \(25^{\circ}\) when they move past the window, as the drawing shows. How fast is the train moving?

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