Two cars leave Nicville travelling in opposite directions. The speed of the faster car is 3 km/h less than...

Question

twice that of the slower car. After 8 hours the cars are 540 KM apart. How fast is each car traveling.

Answer 1

Let v be the speed of the slower car. The slower car covers 8v km in 8 hours. The faster car is traveling (2v-3) km/hr, and covers 8(2v-3) km in 8 hours. The sum of those distances is 540km:

8v + 8(2v - 3) = 540
24v - 24 = 540
24v = 564
v = 23.5 km/hr

The faster car (2v-3) = 2*23.5 - 3 = 44 km/hr

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Now, let's check the answer. At a combined speed of 23.5 + 44 = 67.5 km/hr, the time required to travel 540km is:

540 / 67.5 = 8

Answer 2

let s be the speed of the slower car. Then speed of the faster car is 2s - 3. Since they travel in opposite direction from a same place, after 8 hrs, their distance will add up to 540km

fast car + slow car = 540

distance of faster car:
d = 8(2s - 3)\

distance of slower car
d = 8s

8s + 8(2s - 3) = 540
8s + 16s - 24 = 540
24s = 564
s = 23.5 km/hr

so the speed of the slower car is 23.5 km/hr
speed of the faster car is 2(23.5) - 3 = 44 km/hr

Answer 3

let faster car speed equal vf,
let slower car speed equal vs,
let the distance that the faster car drived be Df
let the distance that the faster car drived be Ds
vf = 2*vs - 3 ...........eq1
Df + Ds = 540.......eq2
the two cars has the same time, so
8 = Df/vs = Ds/vs
=> Df = 8vf, Ds = 8vs
eq2 becomes
8vf + 8vs = 540
dividing eq2 on 8
vf + vs = 67.5 => vf = 67.5 - vs ...eq3
substitute eq3 into eq1
67.5 - vs = 2*vs - 3
70.5 = 3vs
vs = 23.5km/h for the slower car
vf = 67.5 - 23.5 = 44 km/h for the faster car