Two planes leave los angeles at the same time. one heads south to san diego, while the other heads north to san francisco. the san diego plane flies 50 mph slower than the san francisco plane. in 12 hr, 1 half , hr comma the planes are 275 mi apart. what are their speeds?

V1 = 250 mph

V2 = 300 mph

Explanation:

In this case, I will call the plane heading to San Diego as "1" or V1, and the plane heading to San Francisco will be "2" or V2.

Now we know that San Diego is 50 mph slower than San Francisco. So we can actually say that:

V1 = V2 - 50

We know that in half an hour (30 minutes or 0.5 h) both planes are 275 miles apart, this means that the distance is obtained by summing the distance of both planes

If d = V*t then:

d = d1 + d2

replacing the data we have:

275 = (V1*0.5) + (V2*0.5)

Solving for V1:

275 = 0.5V1 + 0.5V2

275 - 0.5V2 = 0.5V1

We also know that V1 = V2 - 50, so if we replace it in the above formula:

0.5(V2 - 50) = 275 - 0.5V2

Then solving for V2:

0.5(V2 - 50) + 0.5V2 = 275

0.5V2 - 25 + 0.5V2 = 275

V2 = 275 + 25

V2 = 300 mph

Now, calculating the value of V1:

V1 = 300 - 50

V1 = 250 mph

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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