Question

At 12 am, train A sets off due west at a fixed speed of s mph. Train B sets off due east from the same station 2 hours later traveling 10 mph faster than train A. The trains are 300 miles apart at 7 am, what is the speed of train A?

Answer 1

Answer: 20.833mph.

Step-by-step explanation:

Using the motion Formula,

speed = distance/time.

This means:

distance = speed * time.

To find the value of "s" in the question, we have to calculate the distance both trains travelled in terms of "s" and sum it to give us 300miles.

For train A,

Speed = S miles per hour

time taken to reach current position= 7hours.

Hence, the distance travelled by train A

Distance A = 7s.

For train B,

Speed= (S + 10)miles per hour.

time taken to reach current position= 5hours (since train B left two hours after the departure of train A)

Distance = speed * time

Distance B = (S + 10) * 5

Distance B = 5s + 50.

Hence, adding both distance to sum up to 300miles, it becomes:

300 = 7s + 5s + 50

300-50 = 12s

250 = 12s

s = 20.833mph.

Answer 2

The speed of train A is 20.83 mph, while that of train B is 30.83 mph

Speed is the ratio of total distance travelled to total time taken. It is given by:

Speed = distance / time

Let d₁ represent the distance covered by train A at 7 am (t = 7 hours) and d₂ represent the distance covered by train B at 7 am (t = 5 hours(7 - 2)).

For train A:

s = d₁ / 7

d₁ = 7s

For train B:

(s + 10 )= d₂ / 5

d₂ = 5s + 50

d₁ + d₂ = 300

7s + 5s + 50 = 300

12s = 250

s = 20.83 mph

s + 10 = 30.83 mph

The speed of train A is 20.83 mph, while that of train B is 30.83 mph

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