Question

Marcia flew her ultra-light plane to a nearby Town against the headwind of 15 km per hour in 2 hours and 20 minutes. The return trip under the same wind conditions took an hour 24 minutes. Find the planes are speed and distance to the nearby town

Answer 1

Speed= 60 km/h

Distance to the nearby town= 105 km

Explanation:

Given: Wind speed= 15 km/h

Time taken while going
`(t_1)`= 2 h 20 minutes.

Time taken while returning
`(t_2)`= 1 h 24 minutes

First we will get the time in uniform unit of hours,

we will convert minutes in hours for
`(t_1)` to get
`2* (20)/(60) = (7)/(3)`

∴
`t_1 = (7)/(3) \ h`

Similarly for
`t_2 = 1* (24)/(60) = (7)/(5) \ h`

∴
`t_2= (7)/(5) \ h`

Now, lets take Plane´s speed as a and distance as d.

While going to the nearby town it is headwind,

∴
`s_1= (a - 15) \ km/h`

While returning it is tailwind

∴
`s_2= (a + 15) \ km/h`

Next, we will use the formula,
`Distance = speed* time`

As distance remain same:
`s_1* t_1 = s_2* t_2`

⇒
`(7)/(3) * (a - 15) = (7)/(5) * (a + 15)`

Now, divide both side by 7

⇒
`(1)/(3) * (a-15) = (1)/(5) * (a+15)`

cross multiplying both side

⇒
`5a-75 = 3a + 45`

∴
`a= 60 \ km/h`

Speed of plane is 60 km/h

Now substituting the value of a in either of equation to get distance.

`d= (7)/(3) * (a-15) = (7)/(3) * (60-15)`

∴ d=
`(7)/(3) * 45 = 105\ km`

105 km is the distance to the nearby town.