Answer:
The slower car travels at 90km/hr.
Step-by-step explanation:
D = total distance both cars cover
t = time passed from starting point to when the cars meet
Va=the speed of the slower car
Vb=the speed of the faster car
DA=the distance travelled by the slower car.
DB=the distance travelled by the faster car.
The sum of the distances travelled by each car equals the total distance between the two starting points.
D = DA + DB
The distance travelled for each car is the speed multiplied by the time elapsed.
DA = VAt
DB = VBt
From the problem we know, VB = VA + 20
Putting all of this together,
D = VAt + (VA + 20)t,
D/t = 2VA + 20
VA = D/2t - 10
plugging in all the values: VA = 90km/hr.
the faster automobile travels at VB = 110km/hr
Answer: the rate of the slower car is 92 kilometers/ hour
Step-by-step explanation:
Initially, the two cars were 1000 kilometres apart before they started travelling towards each other.
Let the speed of the faster car be y km/ h
Let the speed of the slower car be x km/h
One car's rate is 16 kilometers per hour less than the other's. This means that
y = x + 16
x = y - 16
If they meet in 5 hours, it means that both cars have covered a total distance 1000 miles.
Distance = speed × time
Distance travelled by the faster car in 5 hours is
y × 5 = 5y kilometers
Distance travelled by the slower car in 5 hours is
x × 5 = 5x kilometers
Therefore,
5x + 5y = 1000
x + y = 200 - - - - - - - - -1
Substituting y = x + 16 into equation, it becomes
x + x + 16 = 200
2x = 184
x = 184/2 = 92 kilometers/ hour.
y = + 16 + x
y = 16 + 92. = 208 kilometers
