/*! 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 86 A train started from a station a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 86

A train started from a station and, moving with a constant acceleration, covered a distance of \(4 \mathrm{~km},\) finally reaching a speed of 72 \(\mathrm{km} /\) hour. Find the acceleration of the train, and the time taken for the \(4 \mathrm{~km}\)

Short Answer

Expert verified
The acceleration is \(0.05 \text{ m/s}^2\) and the time taken is \(400\) seconds.

Step by step solution

01

Convert Units

First, convert the speed from km/h to m/s for consistency in units. We have a final speed \( v = 72 \text{ km/h} \). To convert this to m/s, use the conversion factor \( \frac{5}{18} \). Thus, \( v = 72 \times \frac{5}{18} = 20 \text{ m/s} \).
02

Identify Known Values

From the problem, identify the known quantities: the final speed \( v = 20 \text{ m/s} \), the initial speed \( u = 0 \text{ m/s} \) since the train started from rest, and the distance covered \( s = 4 \text{ km} = 4000 \text{ m} \).
03

Use the Equation of Motion for Acceleration

We use the second equation of motion: \( v^2 = u^2 + 2as \). Substitute the known values to find the acceleration:\[20^2 = 0^2 + 2a \times 4000\]\[400 = 8000a\]\[a = \frac{400}{8000} = 0.05 \text{ m/s}^2\]Thus, the acceleration of the train is \(0.05 \text{ m/s}^2\).
04

Find the Time Taken

Next, use the first equation of motion \( v = u + at \) to find the time taken. Substitute the known values:\[20 = 0 + 0.05t\]\[t = \frac{20}{0.05} = 400 \text{ s}.\]The time taken for the train to cover the \(4 \text{ km}\) is \(400\) seconds.

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 91影视!

Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause this motion. In our problem, the train's journey from the station to reaching a speed of 72 km/h involves kinematic analysis. We focus on parameters like displacement, velocity, and acceleration:
  • Displacement: This refers to the overall change in position. In our scenario, the train covers a distance of 4 kilometers.
  • Velocity: This is the rate of change of displacement with time. The final velocity of the train is 20 meters per second after conversion from km/h.
  • Acceleration: Here we consider constant acceleration, meaning the train's velocity increases at a steady rate.
Understanding these concepts helps in effectively analyzing problems involving motion, such as the train's acceleration and time to cross the distance.
Equations of Motion
The equations of motion are essential tools in kinematics for solving problems related to moving objects. These equations relate displacement, initial velocity, final velocity, acceleration, and time. In solving the train problem, we utilize two primary equations:
  • The second equation of motion: \[v^2 = u^2 + 2as\] This connects velocity, acceleration, and displacement. Given the train starts from rest, the initial velocity \( u = 0 \). Using the given final speed and distance, we solve for acceleration.
  • The first equation of motion: \[v = u + at\] This formula links velocity with acceleration and time. With known values, calculating the time taken for the train to achieve its speed becomes straightforward.
Having these equations handy simplifies the process of finding one variable when others are known.
Unit Conversion
Unit conversion is crucial in physics to ensure consistency in solving problems. Units must match to utilize equations correctly. In this problem, speed was provided in kilometers per hour, while calculations are done in meters per second.
  • Speed conversion: The formula to convert speed from km/h to m/s is multiplying by the factor \( \frac{5}{18} \). Thus, \( 72 \text{ km/h} \) becomes \( 20 \text{ m/s} \).
  • Distance conversion: The distance needs to be in meters rather than kilometers, hence \( 4 \text{ km} \) converts to \( 4000 \text{ m} \).
Regularly confirming units before calculating ensures that the values fit seamlessly into the equations, minimizing errors and ensuring accurate solutions.
Constant Acceleration
Constant acceleration refers to a scenario where the rate of change of velocity of an object remains the same throughout. In our train example, the train accelerates at a constant rate of \(0.05 \text{ m/s}^2\).
This simplification allows us to use linear equations of motion effectively. With constant acceleration:
  • Velocity increases linearly with time, making the prediction of speed at any time point straightforward using equations of motion.
  • Displacement can be accurately predicted since it correlates quadratically with time, given constant acceleration.
Understanding constant acceleration helps in predicting future motion and understanding past trajectories, a fundamental concept in physics that applies across numerous scenarios.

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

Two old women set out at sunrise and each walked with a constant speed. One went from \(A\) to \(B\), and the other went from \(B\) to \(A\). They met at noon, and continuing without a stop, they arrived respectively at \(B\) at \(4 \mathrm{pm}\) and at \(A\) at \(9 \mathrm{pm} .\) At what time was sunrise on that day? \(\triangle\)

(Average speed of an accelerating car) A typical car (and maybe also a typical train!) does not move with constant acceleration. Starting from a standstill, a car moves through the gears and "accelerates more quickly" in lower gears, when travelling at lower speeds, than it does in higher gears, when travelling at higher speeds. Use this empirical fact to prove that the average speed of a car accelerating from rest is more than half of its final measured speed after the acceleration.

Katya and her friends stand in a circle in such a way that the two neighbours of each child are of the same gender. If there are five boys in the circle, how many girls are there?

[From Paolo dell'Abbaco's Trattato d'aritmetica] "From here to Florence is 60 miles, and there is one who walks it in 8 days [in one direction], another in five days [in the opposite direction]. It is asked: Departing at the same time, in how many days will they meet?"

A paddle-steamer takes five days to travel from St Louis to New Orleans, and takes seven days for the return journey. Assuming that the rate of flow of the current is constant, calculate how long it takes for a raft to drift from St Louis to New Orleans.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation
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.