Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hour less than the other's. if they meet in 4 hours, what is the rate of the slower car?

D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

v₂ = speed of other car = v₁ - 10

t = time taken to meet = 4 h

distance traveled by one car in time "t" + distance traveled by other car in time "t" = D

v₁ t + v₂ t = D

(v₁ + v₂) t = D

inserting the values

(v₁ + v₁ - 10) (4) = 680

v₁ = 90 km/h

rate of slower car is given as

v₂ = v₁ - 10

v₂ = 90 - 10 = 80 km/h

0.863333

86%

hope this

b.

explanation: