You and your friend left a bus terminal at the same time and traveled in opposite directions. your bus was in heavy traffic and had to travel 20 miles per hour slower than your freind's bus. after 3 hours, the buses were 270 miles apart. how fast was each bus going?

The rate at which bus 1 is going is 55 mph

The rate at which bus 1 is going is 35 mph

Explanation:

As per the question:

Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h

Thus  

Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.

R' = R - 20

The distance is given as the product of rate and time:

d = Rt         (1)

Now, the total distance given is 270 miles:

d + d' = 270

Now, using eqn (1):

Rt + R't = 270

3(R + R - 20) = 270

6R = 270 + 60

R = 55 mph

R' = R - 20 = 55 - 20 = 35 mph

speed of the two vehicle are 55 mph and 35 mph

Explanation:

given,

speed of friends vehicle = x mph

speed of your vehicle = (x - 20) mph

when both travel in opposite direction

distance between the two buses = 270 miles

distance = speed × time

270 = 3(x) + 3(x-20)                    

90 = 2 x -20                        

x = 55 mph                    

now, speed of other vehicle is (55-20) = 35 mph

hence, speed of the two vehicle are 55 mph and 35 mph

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

increases it