Question

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C at an average speed of 60 kilometers per hour. If the train did not stop at Station B, what was the average speed at which the train traveled from Station A to C?

(1) The distance that the train traveled from Station A to Station B was 4 times the distance that train traveled from Station B to Station C.
(2) The amount of time it took to the train to travel from Station A to Station B is 3 times the amount of time that it took the train to travel from Station B to Station C.

Answer 1

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Step-by-step explanation:

1)

`v_(ab)` = Speed of train from station A to station B = 80 kmh⁻¹

`d_(ab)` = distance traveled from station A to station B

`t_(ab)` = time of travel between station A to station B

we know that

`Time = (distance)/(speed)`

`t_(ab) = (d_(ab))/(v_(ab)) = (d_(ab))/(80)`

`d_(bc)` = distance traveled from station B to station C

`v_(bc)` = Speed of train from station B to station C = 60 kmh⁻¹

`t_(bc) = (d_(bc))/(v_(bc)) = (d_(bc))/(60)`

Total distance traveled is given as

`d = d_(ab) + d_(bc)`

Total time of travel is given as

`t = t_(ab) + t_(bc)`

Average speed is given as

`v_(avg) = (d)/(t) \\v_(avg) = (d_(ab) + d_(bc))/(t_(ab) + t_(bc))\\v_(avg) = (d_(ab) + d_(bc))/(((d_(ab))/(80) ) + ((d_(bc))/(60) ) )`

Given that :

`d_(ab) = 4 d_(bc)`

So

`v_(avg) = (4 d_(bc) + d_(bc))/(((4 d_(bc))/(80) ) + ((d_(bc))/(60) ) )\\v_(avg) = (4 + 1)/(((4 )/(80) ) + ((1)/(60) ) )\\v_(avg) = 75 kmh^(-1)`

2)

`v_(ab)` = Speed of train from station A to station B = 80 kmh⁻¹

`t_(ab)` = time of travel between station A to station B

`d_(ab)` = distance traveled from station A to station B

we know that

`distance = (speed) (time)`

`d_(ab) = v_(ab) t_(ab)\\d_(ab) = 80 t_(ab)`

`d_(bc)` = distance traveled from station B to station C

`v_(bc)` = Speed of train from station B to station C = 60 kmh⁻¹

`t_(bc)` = time of travel for train from station B to station C

we know that

`distance = (speed) (time)`

`d_(bc) = v_(bc) t_(bc)\\d_(bc) = 60 t_(bc)`

Total distance traveled is given as

`d = d_(ab) + d_(bc)\\d = 80 t_(ab) + 60 t_(bc)`

Total time of travel is given as

`t = t_(ab) + t_(bc)`

Average speed is given as

`v_(avg) = (d)/(t) \\v_(avg) = (d_(ab) + d_(bc))/(t_(ab) + t_(bc))\\v_(avg) = (80 t_(ab) + 60 t_(bc))/(t_(ab) + t_(bc))`

Given that :

`t_(ab) = 3 t_(bc)`

So

`v_(avg) = (80 t_(ab) + 60 t_(bc))/(t_(ab) + t_(bc))\\v_(avg) = (80 (3) t_(bc) + 60 t_(bc))/((3) t_(bc) + t_(bc))\\v_(avg) = ((300) t_(bc))/((4) t_(bc))\\v_(avg) = 75 kmh^(-1)`