Question

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.

(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?

(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?

(c) What was his total displacement relative to the air mass?

Answer 1

a)
`s=35794.2\ m`

b)
`v_w=3.53\ m.s^(-1)`

c)
`s_w=56074.2\ m`

Step-by-step explanation:

Given:

a)

duration of flight,
`t=169* 60=10140\ s`

velocity of flight,
`v=3.53\ m.s^(-1)`

direction of flight,
`45^(\circ)` to the south of east

Now the total displacement:

`s=v.t`

`s=3.53* 10140`

`s=35794.2\ m`

b)

Velocity of air,
`v_a=2\ m.s^(-1)`

When the aircraft encounters a headwind in the opposite direction to the velocity of motion then the speed of the aircraft is lowered with respect to the ground.

But when the speed is observed with respect to the wind the reduced velocity of the aircraft is observed from an opposite moving wind having a magnitude equal to the difference in velocity of the aircraft. This results in no change in the apparent velocity of the aircraft.

Mathematically:

Velocity of the aircraft with respect to the ground:

`v_(g)=v-v_a`

`v_(g)=3.53-2`

`v_g=1.53\ m.s^(-1)`

Now the velocity of the aircraft with respect to the wind:

`v_w=v_g+v_a`

`v_w=1.53+2`

`v_w=3.53\ m.s^(-1)`

c)

Now the total displacement with respect to the wind:

`s_w=v_w.t+v_a.t`

`s_w=3.53* 10140+2* 10140`

`s_w=56074.2\ m`