Does out-of-order makea machine's performance more or less
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Answer:
Here is the C++ program:
#include <iostream> // to use input output functions
#include <cmath> // to use math functions like sqrt()
#include <iomanip> //to use setprecision method
using namespace std; //to access objects like cin cout
int main () { //start of main function
double speedA; //double type variable to store average speed of car A
double speedB; //double type variable to store average speed of car B
int hour; //int type variable to hold hour part of elapsed time
int minutes; //int type variable to hold minutes part of elapsed time
double shortDistance; // double type variable to store the result of shortest distance between car A and B
double distanceA; //stores the distance of carA
double distanceB; //stores the distance of carB
double mins,hours; //used to convert the elapsed time
cout << "Enter average speed of car A: " << endl; //prompt user to enter the average speed of car A
cin >> speedA; //reads the input value of average speed of car A from user
cout << "Enter average speed of car B: " << endl ; //prompt user to enter the average speed of car B
cin >> speedB; //reads the input value of average speed of car A from user
cout << "Enter elapsed time (in hours and minutes, separated by a space): " << endl; //prompts user to enter elapsed time
cin>> hour >> minutes; //reads elapsed time in hours and minutes
mins = hour * 60; //computes the minutes using value of hour
hours = (minutes+mins)/60; //computes hours using minutes and mins
distanceA = speedA * (hours); // computes distance of car A
distanceB = speedB * (hours); //computes distance of car B
shortDistance =sqrt((distanceA * distanceA) + (distanceB * distanceB)); //computes shortest distance using formula √[(distanceA)² + (distanceB)²)]
cout << "The (shortest) distance between the cars is: "<<fixed<<setprecision(2)<<shortDistance;
//display the resultant value of shortDistance up to 2 decimal places
Explanation:
I will explain the program with an examples:
Let us suppose that the average speeds of cars are:
speedA = 70
speedB = 55
Elapsed time in hours and minutes:
hour = 2
minutes = 30
After taking these input values the program control moves to the statement:
mins = hour * 60;
This becomes
mins = 2 * 60
mins = 120
Next
hours = (minutes+mins)/60;
hours = (30 + 120) / 60
= 150/60
hours = 2.5
Now the next two statements compute distance of the cars:
distanceA = speedA * (hours);
this becomes
distanceA = 70 * (2.5)
distanceA = 175
distanceB = speedB * (hours);
distanceB = 55 * (2.5)
distanceB = 137.5
Next the shortest distance between car A and car B is computed:
shortDistance = sqrt((distanceA * distanceA) + (distanceB * distanceB));
shortDistance = sqrt((175 * 175) + (137.5 * 137.5))
= sqrt(30625 + 18906.25)
= sqrt(49531.25)
= 222.556173
shortDistance = 222.56
Hence the output is:
The (shortest) distance between the cars is: 222.56
