Answer sheet – Practice: Muntaha and Eric

1. Muntaha(she/hers) paddles 2.8 m/s [W] from her dock, but there is a current in the lake moving at 1.5 m/s [S]. Find her resultant velocity as observed by a person on shore.

Solution:

$\vec{v_{C W}} = 2.8 \frac{m}{s} [W]$

$\vec{v_{W S}} = 1.5 \frac{m}{s} [S]$

$\vec{v_{C S}} = \vec{v_{C W}} + \vec{v_{W S}}$

Diagram of resulting vector CS with it component vectors: CW = 2.8 m/s & WS 1.5 m/s.

$\vec{v_{C S}} = \vec{v_{C W}} + \vec{v_{W S}}$

$|\vec{v_{C S}}| = \sqrt{{|\vec{v_{C W}}|}^{2} + {|\vec{v_{W S}}|}^{2}}$

$|\vec{v_{C S}}| = \sqrt{{(2.8 \frac{m}{s})}^{2} + {(1.5 \frac{m}{s})}^{2}}$

$|\vec{v_{C S}}| = \sqrt{10.09 \frac{m^{2}}{s^{2}}}$

$|\vec{v_{C S}}| = 3.18 \frac{m}{s}$

$\tan θ = \frac{1.5 m / s}{2.8 m / s}$

$θ = \tan^{- 1} \frac{1.5 m / s}{2.8 m / s}$

$θ = 28^{\circ}$

Her velocity relative to a person on the shore is 3.2 m/s [W28°S]

2. Eric(he/they) wants to fly from Toronto to North Bay, a displacement of 320 km [N]. Eric flies at 120 km/h [N], but there is a wind of 40 km/h blowing from the east.

$\vec{v_{P A}}$ = Velocity of plane relative to the air (airspeed) = 120 km/h [N]
$\vec{v_{A G}}$ = Velocity of air relative to ground (windspeed) = 40 km/h [W]
$\vec{v_{P G}}$ = Velocity of plane relative to ground (groundspeed) = ?

a) What is Eric’s resultant velocity as observed by a person on the ground?

Solution:

$\vec{v_{P G}} = \vec{v_{P A}} + \vec{v_{A G}}$

$|\vec{v_{P G}}| = \sqrt{{|\vec{v_{P A}}|}^{2} + {|\vec{v_{A G}}|}^{2}}$

$|\vec{v_{P G}}| = \sqrt{{(120 \frac{km}{h})}^{2} + {(40 \frac{km}{h})}^{2}}$

$|\vec{v_{P G}}| = 127 \frac{km}{h}$

$\tan θ = \frac{40 km / h}{120 km / h}$

$θ = \tan^{- 1} \frac{40 km / h}{120 km / h}$

$θ = 18^{\circ}$

The answer is 127 km/h [N18°W]

b) If Eric follows this flight path, will he reach the destination?

Solution:

No, the wind carries the plane off course. The plane will be carried west of the destination.

c) Make a sketch to show the path that Eric should have initially followed, in order to fly directly to the desired destination.

Solution:

The resultant velocity, or ground speed, points north from Toronto to North Bay. Its magnitude is unknown. Notice how the vectors representing Eric’s airspeed and the windspeed add tip to tail so their sum results in a resultant velocity that points north.

d) At what angle should he aim the plane, so that he can follow this new path?

Solution:

Using the diagram from c), find the angle by Toronto. This will tell us how far east of north he should aim so that he will actually head straight north.

$\sin θ = \frac{40 km / h}{120 km / h}$

$θ = \sin^{- 1} (0.33)$

$θ = {19.5}^{\circ}$

$[N {19.5}^{\circ} E]$

e) What is Eric's final velocity, if he follows this new path?

Solution:

This means find the new $|\vec{v_{P G}}|$ for the triangle in c). It is the adjacent side of the triangle if you use the angle found in d), the line going north from Toronto to North Bay.

$\cos 1 {9.5}^{\circ} = \frac{|\vec{v_{P G}}|}{120 \frac{km}{h}}$

$|\vec{v_{P G}}| = (\cos 1 {9.5}^{\circ}) (120 \frac{km}{h})$

$|\vec{v_{P G}}| = 113 \frac{km}{h} [N]$

f) How long will it take Eric to get to North Bay, if he follows this new route?

Solution:

$t = \frac{d}{v} = \frac{320 km}{113 km / h} = 2.8 h$