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# Clarify

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An airplane is heading north with an airspeed of 400 km/h. The plane encounters a wind from the east at 100 km/h. What is the magnitude of the resultant ground velocity?

a) 387 km/h

b) 400 km/h

c) 412 km/h

d) 500 km/h

So this got me a bit confused: I did:

√[ (400)^2+(100)^2 ] = 412 km/h

HOWEVER

√[ (400)^2-(100)^2 ] = 387 km/h

Which one is correct? I think it's C though

May 23, 2018

### 1+0 Answers

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Let  the airplane be represented by the  vector     < 0, 400 >

Let the wind be represented  by the vector    < -100, 0 >

The components of the  airplane  vector  are     400 cos 90  = 0     and   400 sin 90  = 400

The components of the wind vector  are   -100 cos 180  = 100    and -100 sin180  = 0

So....the resultant vector  is found as

sqrt [ ( 0 + 100 )^2  + ( 400 + 0 )^2  ] =

sqrt  [ 10000 + 160000 ] =

sqrt [ 170000]  =

412.3 km / h   =  412 km/h  [rounded ]   May 23, 2018