Question

two cars leave city A at the sametime to travel to city B. One car travels at 72km/h and the other travels at 78 km/h. if the slower car arrives 2o mins after the faster car.. how far apart are city A and city B?

pls help me answer dis

Answer 1

u r in grade ????

Answer 2

grade 10

Answer 3

pls help me.

Answer 4

Let A and B be s km apart

Distance covered is velocity*time

Car A is travelling at 72km/h
So it takes car A 72x hours to reach city B

Car B is travelling at 78km/h
So it takes car B 78y hours to reach city B

We know that x = y+1/3, because it takes car A 20 minutes longer (1/3 hours longer)
And they cover the same distance so 72(y+1/3) = 78y
72y + 24 = 78y
y = 4

And the distance is 78*4 = 312km

Answer 5

\[t=\frac{s}{v_1}-\frac{s}{v_2}=s\cdot\frac{v_2-v_1}{v_1v_2}\]
\[s=\frac{v_1v_2}{v_2-v_1}\cdot t=\frac{72\cdot 78}{78-72}\cdot\frac{1}{3}km=312\,km\]

Answer 6

First find the distance between the two cars when the faster one has arrived at city B (which is):
72x20/60 = 24km difference

Then, since the faster car drives 6km/hr faster than the slower one, to find the total time needed to get to city B with the faster car you:

24/6 = 4hrs

Then you multiply this by the speed of the faster car:

4x78 = 312km