/*! 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 55 A truck tractor pulls two traile... [FREE SOLUTION] | 91Ó°ÊÓ

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A truck tractor pulls two trailers, one behind the other, at a constant speed of \(100 \mathrm{~km} / \mathrm{h}\). It takes \(0.600 \mathrm{~s}\) for the big rig to completely pass onto a bridge \(400 \mathrm{~m}\) long. For what duration of time is all or part of the truck-trailer combination on the bridge?

Short Answer

Expert verified
The duration of time all or part of the truck-trailer combination is on the bridge is approximately \(14.99\) seconds.

Step by step solution

01

- Convert speed to m/s

The given speed is \(100 \mathrm{~km} / \mathrm{~h}\). To convert this to m/s, multiply by \(1000 / 3600\) because there are 1000m in a km and 3600 seconds in an hour. This gives \(100 \times (1000 / 3600) = 27.8 \mathrm{~m/s}\).
02

- Calculate the time to cross the bridge

Now we calculate the time it takes for the truck-trailer combination to cross the bridge with the speed we calculated in step 1. We use the formula for time which is \(time = distance / speed\), substituting the given values, the time it takes to cross the bridge is \(400 / 27.8 = 14.39 \mathrm{~s}\).
03

- Calculate total time

Having calculated the time it takes to cross the bridge, the total time that the truck-trailer combination is on the bridge also includes the time it takes to completely pass onto the bridge, which is given as \(0.600 \mathrm{~s}\). Adding these two durations gives the total time on the bridge: \(14.39 + 0.600 = 14.99 \mathrm{~s}\).

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Key Concepts

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

Constant Speed Motion
Understanding the concept of constant speed motion is fundamental in physics. It's a condition when an object moves at a uniform speed, neither speeding up nor slowing down. In such a scenario, we assume that the forces acting on the object such as friction or air resistance are negligible, or perfectly balanced by the driving force, resulting in no acceleration.

For example, our truck tractor pulling trailers in the exercise is said to move at a constant speed of 100 km/h. This implies that the time it takes to travel any given distance can be calculated with a simple formula: time equals distance divided by speed. This is what makes the problems involving constant speed relatively straightforward to solve, as long as all the variables are known.

In real-world scenarios, like the one presented, we can use constant speed motion to determine travel time over a bridge, or the time taken for a car to traverse a tunnel. Such problems are not only common in textbooks but also in practical life, such as route planning and logistics.
Unit Conversion
Whenever you're solving a physics problem, it's crucial to pay close attention to the units.

Why Convert Units?

In our daily lives, we may encounter various units of measurement. To work effectively with these different units, especially within equations, we need to convert them into a consistent system, like the International System of Units (SI).

How to Convert Units

In the truck-trailer exercise, the speed is given in kilometers per hour (km/h) but we need to work in meters per second (m/s) since most physics equations use the SI units. To convert km/h to m/s, we divide the value by 3.6 (since 1 km equals 1000 meters and 1 hour contains 3600 seconds). This is a significant step because using different units could cause errors in calculations, leading to incorrect solutions to the problems.

Unit conversion is a transferrable skill that is used in various fields, not just in physics, making it an essential part of a student's skill set in science and engineering.
Motion Equations
The backbone of kinematics, the study of motion, lies within its equations, known as the motion equations. These equations relate displacement, velocity, acceleration, and time. Although the truck-trailer problem only requires a simple distance-over-time calculation due to constant speed, knowing the motion equations is key for more complex problems involving acceleration.

The basic formula, as applied in our truck-trailer example, is \(time = \frac{distance}{speed}\), which is derived from the definition of speed as distance per unit of time. This equation assumes that the speed is constant over the distance covered. However, when speed changes over time, we need to use other equations from the set of motion equations which account for acceleration.

These equations allow us to predict future positions and velocities of moving objects, analyze the motions to deduce the forces involved, and solve a vast array of problems. From designing roller coasters to planning space missions, motion equations are applied in many different areas within physics and beyond.

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Most popular questions from this chapter

A parachutist with a camera descends in free fall at a speed of \(10 \mathrm{~m} / \mathrm{s}\). The parachutist releases the camera at an altitude of \(50 \mathrm{~m}\). (a) How long does it take the camera to reach the ground? (b) What is the velocity of the camcra just before it hits the ground?

An attacker at the base of a castle wall \(3.65 \mathrm{~m}\) high throws a rock straight up with speed \(7.40 \mathrm{~m} / \mathrm{s}\) at a height of \(1.55 \mathrm{~m}\) above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of \(7.40 \mathrm{~m} / \mathrm{s}\) and moving between the same two points. (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at \(71.5 \mathrm{~m} / \mathrm{s}\). The driver of the Thunderbird realizes that she must make a pit stop, and she smoothly slows to a stop over a distance of \(250 \mathrm{~m}\). She spends \(5.00 \mathrm{~s}\) in the pit and then accelerates out, reaching her previous speed of \(71.5 \mathrm{~m} / \mathrm{s}\) after a distance of \(350 \mathrm{~m}\). At this point, how far has the Thunderbird fallen behind the Mercedes Benz. which has continued at a constant speed?

Two students are on a balcony \(19.6 \mathrm{~m}\) above the street. One student throws a ball vertically downward at \(14.7 \mathrm{~m} / \mathrm{s} ;\) at the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. (a) What is the difference in the two balls' time in the air? (b) What is the velocity of each ball as it strikes the ground? (c) How far apart are the balls \(0.800 \mathrm{~s}\) after they are thrown?

A certain cable car in San Francisco can stop in \(10 \mathrm{~s}\) when traveling at maximum speed. On one oceasion, the driver sees a dog a distance \(d \mathrm{~m}\) in front of the car and slams on the brakes instantly. The car reaches the dog \(8.0 \mathrm{~s}\) later, and the dog jumps off the track just in time. If the car travels \(4.0 \mathrm{~m}\) beyond the position of the dog before coming to a stop, how far was the car from the dog? (Hint: You will need three equations.)

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