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Going from Boston to SF 4344/6.33 = 685.895 km/h

Going from SF to Boston 4344/5.33 = 814.5 km/h

814.5 - 685.895 = 128.605

128.605/2 = 64.3026 km/h

Let the velocity of the plane relative to the ground from Boston to San Francisco be $v_w$ and the velocity relative to the ground on the reverse trip be $v_e$ . Using $d=vt$ , we then have

$v_w=\frac{4344~km} {6.33~hr}=686.26~\mbox{km/hr}$ , $v_e=\frac{4344~km} {5.33~hr}=815.01~\mbox{km/hr}$

We can re-express these velocities in terms of the speed $v$ of the plane relative to the air and the speed $v_J$ of the jet stream, $v_w=v-v_J$ , $v_e=v+v_J$ . We can then solve these two equations for $v$ and $v_J$ .

For some people who are confused on 6.33 hr, you can try to look at it in a different way: 6 hours and 20 minutes = 360 minutes + 20 minutes = 380 minutes. 5 hours and 20 minutes = 300 minutes + 20 minutes = 320 minutes The distance from BOS to SF is 4344 km, and this is based on per hour. We have to convert either minute to hour or vise versa. In this case, we'll convert the distance to per minute: 4344 x 60 (mins) = 260640 km/m Now, we can get the velocity, $\frac{260640 km/m}{380 m}$ = 685.9 km $\frac{260640 km/m}{320 m}$ = 814.5 km

Then subtract the two speeds to get the difference between BOS-SF and SF-BOS. The difference is around 128.61 km. Divide that by 2 and you will get something around 64. Why divide? Because you want to know the speed for one way.

If $v_j$ is the jet velocity, $v_w$ is the wind velocity, and the distance to travel is d, with small time $t_1$ and long time $t_2$ :

$v_j + v_w = \frac{d}{t_1}$ ; $v_j - v_w = \frac{d}{t_2}$

Subtracting the two equations, we arrive at:

$2*v_w = d*(\frac{1}{t_1} - \frac{1}{t_2})$ Since 20 minutes is 1/3 of an hour, we can see that $t_1$ is 16/3, and $t_2$ is 19/3. This leads to the elegant result of:

$v_w = d(\frac{9}{608})$

Substituting in d and $v_w$ resolves to 64.30.

Let A = airline speed and J = Jetstream speed. Then 4344/(A-J) = 6.33 and 4344/(A+J) = 5.33. Two equations, 2 unknowns. Solve for J = 64.38

• Time Boston to SF = 6.33 hours ie $\frac{(60*6)+20}{60}$ to convert time into hours as a decimal
• Time SF to Boston = 5.33 hours in the same way

• Distance = 4344 km

• Airplane speed = $\frac{distance}{time taken}$ this is the airplane speed not jetstream

• Airplane speed SF to Boston = $\frac{4344}{6.33}$ = 814.5 km/hour

• Airplane speed Boston to SF = $\frac{4344}{6.33}$ = 685.89 km/hour

• Speed Boston to SF = (Airplane constant speed) - (jetstream speed)

• Speed SF to Boston = (Airplane constant speed) + (jetstream speed)

• To make it easier we can call jetstream speed x

• The difference between the speeds is (Speed SF to Boston) - (Speed Boston to SF) = 685.89 - 814.5 = 128.61

• So (constant + x) - (constant - x) = 128.61

• We then cancel out the equation:

• Constant - constant cancels each other out
• so x - (-x) = 128.61
• minus a minus is the same as plus

• so x+x = 128.61

• So 2x = 128.61 ie 2x jetstream speed = 128.61

• So x = 128.61/2 = 64.3

For those people who got 128km/h, and are confused why everybody (and the correct solution) involves dividing by 2, i'll explain this in a different way... Imagine if there is no wind, what would be the speed of the plane? Well, it would have to be between 815 and 686 km/h. To be more precise, take the average of those two speeds. So the plane speed would be 751 km/h if there's zero wind. So therefore the wind speed must be 64 km/h, as the plane travels 64km/h faster with the wind and 64 km/h slower against the wind.

$5.33(v_{\text{plane}} + v_{\text{air}})= 4344 = 6.33(v_{\text{plane}} - v_{\text{air}})\\ 5.33 v_{\text{plane}} + 5.33 v_{\text{air}} = 6.33 v_{\text{plane}} - 6.33 v_{\text{air}}\\ v_{\text{plane}} = 11.66 v_{\text{air}}\\ 5.33(11.66 v_{\text{air}} + v_{\text{air}}) = 4344\\ 12.66 v_{\text{air}} = \frac{4344}{5.33}\\ v_{\text{air}} = \frac{4344}{12.66 \cdot 5.33} = 64.37672...\approx 64.4 \text{ km/h}\\$

6h20m = 19/3 h

5h20m = 16/3 h

Speed of plane = A

Speed of jet stream = J

(A+J)*16/3 = 4344

(A-J)*19/3 = 4344

Rearrange to get:

A+J = 3*4344/16

A-J = 3*4344/19

Subtract the two equations to eliminate A:

2J = 3*4344(1/16 - 1/19)

J = 3*2172(1/16 - 1/19)

J = 64.3026

Why did the "official" answer say 64.38, when the information provided leads every correct method to calculate 64.30 km/h?

B = Boston S = San Fran

6Hr 20 = $\frac{19}{3}$ hrs 5Hr20 = $\frac{16}{3}$ hrs

w = wind speed towards B v = plane speed

Travel in direction BS (into headwind so relative plane speed is slower):

Dist = Speed * Time

4344 = (v-w)* $\frac{19}{3}$

$\frac{4344*3}{19}$ = (v-w) ----equ(1)

Travel in direction SB (with tailwind so relative plane speed is faster) Dist = Speed * Time

4344 = (v+w)* $\frac{16}{3}$

$\frac{4344*3}{16}$ = (v+w) ----equ(2)

To get v use equ(1) + equ(2)

To get w equ(2) - equ(1):

$\frac{4344*3}{16}$ - $\frac{4344*3}{19}$ = 2w

w = $\frac{4344*3}{2}$ *( $\frac{1}{16}$ - $\frac{1}{19}$ ) = $\frac{4887}{76}$ = 64.30263...

w = 64.30 KPH (not 64.38 that uses imprecise time values)

Let... Vp=the average velocity of the plane sj=the average velocity of the jet stream

First we calculate the average velocity of the plane heading from San Francisco to Boston. Vp = Δs/t= 4344/5.33 = 815.009 km/hr

From Boston to San Francisco the velocity of the plane is combined with the velocity of the jet stream. Knowing this info we can write the following expression.

Vp + Vj = 815.009

Using the same logic above, we can calculate the speed of the plane on its return to San Francisco from Boston. Making sure that the signs are flipped due to the change in direction. V = Δs/t = -4344/6.33 = -686.255 km/hr

Next we need to consider that the two velocity vectors of the plane on its return trip and the jet stream. They are opposite directions and will have different signs, so we are actually subtracting the two. Knowing this we can derive the following equation...

Vp - Vj = -686.255

Solve the two equations for Vp and set them equal...

Vp = -686.255 +Vj Vp = 815.009 - Vj

815.009 - Vj = -686.255 + Vj 815.009 = -686.255 + 2Vj 128.7534 = 2Vj 64.37 = Vj

The actual velocity of the plane remains constant (as power configurations on airplane will be constant).

When it's going from San Francisco to Boston, the speed of the plane is aided by the speed of the jet stream, and the vice-versa on the trip from Boston to San Francisco.

Thus,

$\frac{6344}{5.33}$ - $v$ = $\frac{6344}{6.33}$ + $v$

Which, gives v = 64.30 (approximately)

The average speed of the plane on the two legs is 750.94 km/hr. This is the plane's ground speed in the absence of wind. If the plane had flown for 6.33 hours at 750.94 km/hr, it would have gone an extra 407.77 km, which is the distance travelled by a particle in the air mass of the jet stream. It travelled that distance in 6.33 hrs, so its speed is 64.38 km/hr.

AIRCRAFT GROUNDSPEEDS: 4344 / 6.33 = 815 KM/h SF -> Bos 4344 / 5.33 = 686.26 KM/h Bos ->SF

Find the difference between these groundspeeds; 815 - 686.26 = 128.74 KM/h

One aircraft was sped up by the jetstream, the other slowed down by it. So the jetstream made an impact on the difference in speeds TWICE. Therefore divide the difference in speed we found above by two;

128.74 / 2 = 64.37 is the speed of the jetstream.

We can check the answer;

If groundspeed SF -> Bos was 815Km/h and the jetstream was as we have calculated, 64.37 km/h, then the aircraft AIRSPEED must have been 815 - 64.37 = 750.63km/h

Compare this with Bos -> SF; 686.26 + 64.37 = 750.63km/h...

The airspeeds match.

Back to the original question posed to us...”the difference in flight times is due to the jetstream” meaning that the aircraft AIRSPEEDs was assumed constant in either direction/flight.

The difference between airspeed and groundspeed is like this;

groundspeed: you are waiting on an airport bench seat and you watch someone move by while walking along a moving travelator. Walking with the travelator they appear to move rapidly relative to you. If that person turned 180 and tried to fight the movement of the travelator (against the flow of other passengers also!) they would appear to move slowly relative to you.

Airspeed: instead of waiting on the airport bench, you are now another passenger standing still, riding the travelator. Someone walking towards you (the wrong way) on the same travelator, or away from you moves at the same rate relative to you. You are now in their frame of reference. Or.... flowing along with their jetstream.

In one direction, the wind is helping, in the other it is hindering.

Vplane + Vwind = 4344km/5hr20min

Vplane - Vwind = 4344km/6hr20min

Subtract the 2 equations to get rid of Vplane

2Vwind = (4344km/5hr20min - 4344km/6hr20min)

Convert 20 min to hours for both 20/60 = 1/3

5+1/3 = 16/3 and 6+1/3 = 19/3

Vwind = (1/2)(4344/(16/3) - 4344/(19/3))

normalizing by multiplying by 19*16

Vwind = (1/2)(4344)(3)(19-16)/(19*16) km/hr

Vwind = 64.30 km/hr, not sure why solution says 64.38 km/hr

On the onward journey, the wind was opposing the travel

$v_1 - v_2 = D/t_1$

and during the return journey, wind was aiding

$v_1 + v_2 = D/t_2$

Solving for $v_2$

$v_2 = \frac{D}{2} (\frac{1}{t_2} - \frac{1}{t_1})$

$v_2 = \frac{4344}{2} (\frac{1}{16/3} - \frac{1}{ 19/3}) = 64.3026$

Let v be the speed of the aeroplane and j be the speed of the jet stream.

Then (v-j)(6h20m) = 4344 . . . (A)

and (v+j)(5h20m) = 4344 . . . (B)

It follows that (v-j)(6h20m) = (v+j)(5h20m)

or (v-j)( $\frac{19}{3}$ h) = (v+j)( $\frac{16}{3}$ h)

or 19(v-j) = 16(v+j) which, on simplification, gives 35j = 3v

or j = $\frac{3v}{35}$ . . . (C)

Substituting equation (C) into equation (B),

(v+j)( $\frac{16}{3}$ ) = 4344

we have (v + $\frac{3v}{35}$ ) = 4344( $\frac{3}{16}$ )

Therefore v = 750.197368421 km/h

and j = $\frac{3v}{35}$ = 64.3026315789 km/h

Averaging the jet's speed removes affect of the wind. The wind speed is the absolute difference between the actual speed, in either direction, and the average speed.

$s_{w}$ = $\frac{d}{t_{w}}$ = $\frac{4344}{6.33}$ = 686.26 km/h

$s_{e}$ = $\frac{d}{t_{e}}$ = $\frac{4344}{5.33}$ = 815.01 km/h

a = $\frac{s_{w}+s_{e}}{2}$ = 750.63 km/h

w = $s_{e}$ - a = 815.01 - 750.63 = 64.38 km/h

or

a = $\frac{d}{2}$ ( $\frac{1}{t_{w}}$ + $\frac{1}{t_{e}}$ )

w = d( $\frac{1}{t_{e}}$ - $\frac{1}{2}$ ( $\frac{1}{t_{w}}$ + $\frac{1}{t_{e}}$ ))

Find the speed to/from Boston to San Fran. Calculate their difference then half.