What is the resultant velocity of the?

Question

1. A boat heads directly across a river 41 m wide at 3.8 m/s. The current is flowing downstream at 2.2 m/s.

a. What is the resultant velocity of the boat?

b. How much time does it take the boat to cross the river?

c. How far downstream is the boat when it reaches the other side?

4. A 42-km/h wind blows toward 215 degree, while a plane heads toward 125 degree at 152 km/h. What is the resultant velocity of the plane?

Answer 1

♠the velocity of the boat is vector w=(u, v), where u=2.2m/s, v=3.8m/s;

resultant speed of the boat is  (Pythagoras)

(a) |w|=√(u^2 +v^2) =√(2.2^2 +3.8^2) =4.39 m/s;

♣here we shouldn’t take u=2.2m/s into account;

(b)therefore t=s/v = 41/3.8 =10.79s;

♦during this very time t=10.79s the river will carry the boat

(c) h=u*t =23.74 meters downstream;  

♥ the velocity of the plane is a vector w=u +v, where

u=(cos(215), sin(215))*42km/h, v=(cos(125), sin(125))*152km/h;

w=(cos(215)*42 +cos(125)*152, sin(215)*42 +sin(125)*152) =

=(-121.59, 100.42)km/h;

speed is |w| =√((-121.59)^2 +100.42^2) = 157.70km/h;

Answer 2

<< What is the resultant velocity of the boat? >>

R^2 = A^2 + B^2

where

A = speed of current = 2.2 m/sec.

B = speed of boat = 3.8 m/sec.

R = resultant velocity of boat

Substituting values,

R^2 = 2.2^2 + 3.8^2

and solving for R,

R = 4.39 m/sec.

<< How much time does it take the boat to cross the river? >>

The angle that the resulant velocity vector makes (with respect to the vertical) is equal to

tan C = 2.2/3.8  = 0.5789

C = arc tan 0.5789 = 30.07 degrees

Effectively,  

the distance travelled by the boat while crossing = 41/cos 30.07 = 47.38 m

Therefore, time for boat to cross the river = 47.38/4.39 = 10.79 sc.

  

<< How far downstream is the boat when it reaches the other side? >>

Let

X = distance downstream travelled by the boat

X/41 = tan 30.07

X = 41(tan 30.07)

X = 23.74 m

<< A 42-km/h wind blows toward 215 degree, while a plane heads toward 125 degree at 152 km/h. What is the resultant velocity of the plane? >>

Use the Law of Cosines to solve this problem, i.e.,

A^2 = B^2 + C^2 - 2(B)(C)(cos a)

where

A = resultant velocity of plane

B = wind speed = 42 kph (given)

C = plane speed = 152 kph (given)

a = angle between the plane and wind vectors = 215 - 125 = 90 degrees

Since the angle "a" = 90 degrees, then the effective formula simply becomes the Pythagorean theorem, i.e.,

A^2 = B^2 + C^2

and substituting values,

A^2 = 42^2 + 152^2

and solving for A,

A = 157.50 kph