Trains A and B, 200 km apart on the same straight track, travel at speeds of 50 km/hr and 65 km/hr respectively. At the end of 1 hour, the distance between the trains could not

Ok, at the beginning the distance between the trains is 200km:

IPa(0s) - P(0s)I = 200km.

where Pa is the position of train A, and Pb is the position of train B.

Now, the speed of train A is:

Sa = 50km/h

Sb = 65km/h

Now, remember the relation:

Distance = Speed*time.

So, in one hour, the displacement of train A is:

Distance = 50km/h*1h = 50km

The displacement of train B is:

distance  = 65km/h*1h = 65km.

Now, if at the beginning train A is 200km ahead of train B, then we have:

Pa - Pb = (Pa(0s) + 50km) - (Pb(0s) + 65km)

= (Pa(0s)  - Pb(0s)) + (50km - 65km) = 200km - 15km = 185km.

Now, if train B was 200km ahead of train A, we have:

Pb - Pa = (Pb(0s) + 65km) - (Pa(0s) - 50km) =

(Pb(0s) - Pa(0s)) + (65km - 50km) = 200km + 15km = 215km

answer: the first graph

step-by-step explanation:

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation: