/*! 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 84 A rocket-driven sled running on ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rocket-driven sled running on a straight, level track is used to investigate the effects of large accelerations on humans. One such sled can attain a speed of \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\), starting from rest. Find (a) the acceleration (assumed constant) in terms of \(g\) and (b) the distance traveled.

Short Answer

Expert verified
(a) Acceleration is about \(25.17g\). (b) Distance traveled is \(400 \text{ m}\).

Step by step solution

01

Convert Speed to Meters Per Second

First, convert the given final speed from kilometers per hour to meters per second. Use the conversion factor\[ 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} \]Thus, the speed in meters per second is:\[ v = 1600 \text{ km/h} \times \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{1600 \times 1000}{3600} \text{ m/s} = \frac{1600 \times 10}{36} = 444.44 \text{ m/s} \]
02

Calculate Acceleration

Since the rocket starts from rest, its initial velocity \( u = 0 \text{ m/s} \). The final velocity \( v = 444.44 \text{ m/s} \), and the time \( t = 1.8 \text{ s} \). The formula for acceleration \( a \) is:\[ a = \frac{v - u}{t} = \frac{444.44 - 0}{1.8} = 246.91 \text{ m/s}^2 \]
03

Express Acceleration in Terms of g

Gravitational acceleration \( g \) is approximately \( 9.81 \text{ m/s}^2 \). To express acceleration \( a \) in terms of \( g \), divide by \( 9.81 \text{ m/s}^2 \):\[ \text{Acceleration in terms of } g = \frac{246.91}{9.81} = 25.17g \]
04

Calculate Distance Traveled

Use the formula for distance traveled with constant acceleration:\[ s = ut + \frac{1}{2}at^2 \]Since the sled starts from rest, \( u = 0 \), so:\[ s = \frac{1}{2} \times 246.91 \times (1.8)^2 \]\[ s = \frac{1}{2} \times 246.91 \times 3.24 = 399.99 \text{ m} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Acceleration
Acceleration refers to how quickly an object changes its velocity. It is a measure of the rate of change of velocity over time. In physics, acceleration is often denoted by the letter "a". When you hear about acceleration in everyday life, it's likely you're considering linear acceleration, which occurs in a straight line.

In the sled problem, the acceleration is calculated given that the initial velocity is zero when starting from rest. We find this using the formula:
  • Acceleration, \( a = \frac{v - u}{t} \)
  • where \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is the time taken.
  • In this case: \( a = \frac{444.44 - 0}{1.8} \) m/s²
Understanding acceleration is crucial in physics problem solving because it affects how quickly or slowly an object will reach a certain velocity. In this sled example, it measures the intensity of change of speed over a short period.
Velocity Conversion
Velocity conversion is an essential step in physics problems involving measurements from different unit systems. In our exercise, the initial speed given was in kilometers per hour (km/h), but we needed it in meters per second (m/s) to correctly apply kinematic equations.

The conversion factor is:
  • \( 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} \)
  • Therefore, a speed of 1600 km/h converts to \( 444.44 \text{ m/s} \) as follows:
  • \( v = 1600 \times \frac{1000}{3600} \)
Converting velocities appropriately ensures consistency across calculations and aligns with the standard SI unit system. It is a simple yet critical step, ensuring you are working within uniform measurement requirements, facilitating accurate physics problem solving.
Gravitational Acceleration
Gravitational acceleration is the acceleration of an object due to the force of gravity. On Earth, this is approximately \( 9.81 \text{ m/s}^2 \), and it's denoted by \( g \). When dealing with high acceleration scenarios, like our rocket-driven sled, it's insightful to express acceleration in terms of \( g \) for better comprehension.

We calculate this by dividing the calculated acceleration by gravitational acceleration:
  • \( \text{Acceleration in terms of } g = \frac{246.91}{9.81} \).
  • This results in approximately \( 25.17g \).
This means the sled experiences a force over 25 times stronger than the gravitational pull on Earth, highlighting the extreme conditions the experiment subjects are exposed to. Converting to "g"-forces helps visualize and communicate the intensity of such conditions for both educational and applied physics purposes.
Distance Traveled
Distance traveled regards how far an object moves during a period of acceleration. In context, calculating this on our sled track lets us know how long the sled runs before reaching its final speed.

The kinematic equation used here is:
  • \( s = ut + \frac{1}{2}at^2 \)
  • where \( s \) is distance, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
  • Given that initial velocity \( u = 0 \), this simplifies to \( s = \frac{1}{2}at^2 \).
  • For the sled, \( s = \frac{1}{2} \times 246.91 \times (1.8)^2 = 399.99 \text{ m} \).
Calculating distance traveled is vital for understanding the scope of motion within a given time and under given acceleration circumstances. This not only helps in physics problem solving but also in real-world applications like designing and testing vehicle tracks.
Kinematics
Kinematics is the branch of physics that describes the motion of points, bodies, and systems without considering the forces that caused the motion. It is fundamental to solving numerous physics problems, such as our sled scenario, as it combines essential variables like velocity, acceleration, time, and distance.

The core kinematic equations drive the analysis of motion:
  • \( v = u + at \)
  • \( s = ut + \frac{1}{2}at^2 \)
  • \( v^2 = u^2 + 2as \)
Applying these equations involves understanding how each variable affects the others. In the case of the sled, knowing any three of these allows us to solve for the fourth. Kinematics helps predict how different settings will influence an object's movement, making it a versatile and powerful tool in physics problem solving, ensuring comprehensive analyses from simple scenarios to complex systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A jumbo jet must reach a speed of \(360 \mathrm{~km} / \mathrm{h}\) on the runway for takeoff. What is the lowest constant acceleration needed for takeoff from a \(1.80 \mathrm{~km}\) runway?

On a dry road, a car with good tires may be able to brake with a constant deceleration of \(4.92 \mathrm{~m} / \mathrm{s}^{2} .\) (a) How long does such a car, initially traveling at \(24.6 \mathrm{~m} / \mathrm{s}\), take to stop? (b) How far does it travel in this time? (c) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for the deceleration.

A motorcyclist who is moving along an \(x\) axis directed toward the east has an acceleration given by \(a=(6.1-1.2 t) \mathrm{m} / \mathrm{s}^{2}\) for \(0 \leq t \leq 6.0 \mathrm{~s}\). At \(t=0\), the velocity and position of the cyclist are \(2.7 \mathrm{~m} / \mathrm{s}\) and \(7.3 \mathrm{~m} .\) (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between \(t=0\) and \(6.0 \mathrm{~s} ?\)

Two particles move along an \(x\) axis. The position of particle 1 is given by \(x=6.00 t^{2}+3.00 t+2.00\) (in meters and seconds); the acceleration of particle 2 is given by \(a=-8.00 t\) (in meters per second squared and seconds) and, at \(t=0\), its velocity is \(20 \mathrm{~m} / \mathrm{s}\). When the velocities of the particles match, what is their velocity?

A ball of moist clay falls \(15.0 \mathrm{~m}\) to the ground. It is in contact with the ground for \(20.0 \mathrm{~ms}\) before stopping. (a) What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? (Treat the ball as a particle.) (b) Is the average acceleration up or down?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.