Two cars leave a city headed in the same direction

Question

Two cars leave a city headed in the same direction. After 4 hours, faster car is 40 miles ahead of the slower car. The slower car has traveled 168 miles. Find the speed of the 2 cars.

Answer 1

Hi Cory,

 

Let F = the speed of the faster car

     S = the speed of the slower car

 

After 4 hours,

• The slower car traveled 168 miles. Its speed can be determined by dividing the distance traveled by the time

             Speed = Distance/time

                       = 168/4

                       42 miles per hour.

• The faster car traveled 40 miles further than the slower car. So, the faster car has travelled for 208 miles (168+40). Therefore, its speed can be determined as follows

              Speed = Distance/time

                        = 208/4

                        52 miles per hour.

 

Hence, the speed of the faster car is 52 miles per hour while the slower car travelled at a speed of 42 miles per hour.

 

I hope it helps!

A.

 

Answer 2

Car a is traveling 168mi/4hr which is 42mph. Car B is traveling 208mi/4hr which is 52mph.