Question

Luke is swimming in a river with a current flowing west to east at a constant speed of 3.5 meters/second. Assume that Luke's water speed is a constant 5 meters/second. If he tries to swim 60° east of north, his resultant speed is 2.63 6.58 7.40 8.22 8.52 meters/second and his direction angle is 17.70 35.82 71.63 103.01 144.18 °.

Answer 1

Decompose each velocity in its i and j - coordinates.


1) River's velocity


3.5 m/s, west fo east = 3.5i + 0j


2) Luke's velocity

5m/s, 60° east of north

5* sin(60°) i + 5*cos(60°) j = 4.33 i + 2.5 j


3) Resultant velocity

[3.5i + 0j] + [4.33i + 2.5j] = 7.83i + 2.5j


resultant speed = √[ (7.83)^2 + (2.5)^2 ] = 8.22 m/s


angle = arctan(2.5 / 7.83) = 17.7° north of east (equal to 72.3° east of north).

Answer 2

Answer:

The resultant speed is 8.22 m/s.

The direction angle is 17.74°, or just 17.7°

Step-by-step explanation:

In this problem we have to operate vector, specifically we have to sum and find rectangular components. Remember that vectors are compound by rectangular components, one horizontal (x) and one vertical (y). So, when we sum or subtract vector, we can do it between similar coordinates, that means we can't sum horizontal with vertical coordinates.

So, in the first movement, we have to sum to horizontal vector, they don't have vertical component because the problem says that Luke is traveling from East to West, without an angle of direction.

Current's speed: $(3.5i)m/s$

Notice that the river only has horizontal component.

Then, we have to calculate the components of the vector in the second movement, because it has a different direction of 60°. To calculate components we use:

For vertical component: $v_{y}=v.sen\alpha$

For horizontal component: $v_{x}=v.cos\alpha$

In this case, $v=5;\alpha =60\°$

So, $v_{x}=5.sen60=4.33i$ and $v_{y}=5.cos60=2.5j$

Then, to calculate the total speed, we just sum like terms, horizontal with horizontal and vertical with vertical:

$V=(3.5+4.33)i+(0+2.5)j\\V=7.83i+2.5j$

Now, to calculate the module of this velocity we use:

$|V|=\sqrt{x^{2}+y^{2}}=\sqrt{(7.83)^{2}+(2.5)^{2} } \\|V|=\sqrt{61.31+6.25} =8.22$

To calculate the direction we use: $\alpha =tan^{-1}(\frac{y}{x})$

$\alpha =tan^{-1}(\frac{2.5}{7.83})\\\alpha=tan^{-1} (0.32)=17.74\°$

Therefore, The resultant speed is 8.22 m/s. And, the direction angle is 17.74°, or just 17.7°