An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown distance, x km away. It travels straight to the fire and back, averaging 240 km/h for the outward trip and 320 km/h for the return trip. If the plane was away for 35 minutes, find the distance, x km.

Is there supposed to be some sort of simultaneous equation involved, with the use of speed = d/t or am I supposed to sub all of the information into the one equation. Also, not sure if I need to convert 35mins into hours or not?

Question

An aircraft, used for fire spotting, flies from its base to locate a fire at an unknown distance, x km away. It travels straight to the fire and back, averaging 240 km/h for the outward trip and 320 km/h for the return trip. If the plane was away for 35 minutes, find the distance, x km.

Is there supposed to be some sort of simultaneous equation involved, with the use of speed = d/t or am I supposed to sub all of the information into the one equation. Also, not sure if I need to convert 35mins into hours or not?

astrophysics · Answer

Yes, convert all the units so they are consistent, and always start by making a diagram. |dw:1451012034001:dw| you have the right idea, we are using v = d/t for the round trip. 
$v_1 = 240 km/h$ and $v_2 = 320 km/h$ and we know that the distance $x_1~~ \text{and}~~ x_2$ are the same so we have $x_1 = x_2 = x$ and the time $t_1+t_2 = 35 \min$ since we know the plane was away for 35 minutes. I think this should be enough information to get your started and figure it out.