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