Answer:
v = 0.78 m/sec at 21.61° towards the east of north
Explanation:
Given a 19.3 kg dog is running northward at 2.98 m/s , while a 6.26 kg cat is running eastward at 3.64 m/s and the mass of the owner is 78.7 kg.
Given m(dog) = 19.3 kg and m(cat) = 6.26 kg
v(dog) = [tex]2.98\vec{j}[/tex] m/sec
v(cat) = [tex]3.64\vec{i}[/tex] m/sec
Given the [tex]\vec{P}(owner)=\vec{P}(dog)+\vec{P}(cat)[/tex]
Let the velocity of the owner be [tex]\vec{v}[/tex]
We know that [tex]\vec{P}=m\vec{v}[/tex]
[tex]m(owner)\times \vec{v}=m(dog)\times \vec{v(dog)}+m(cat)\times \vec{v(cat)}[/tex]
[tex]78.7\times \vec{v}=19.3\times 2.98\vec{j}+6.26\times 3.64\vec{i}[/tex]
[tex]\vec{v}=\frac{1}{78.7}(57.5\vec{j}+22.78\vec{i})[/tex]
[tex]v=\sqrt{(\frac{57.5}{78.7})^2+(\frac{22.78}{78.7})^2 }[/tex]
v=0.78 m/sec
for direction
Tan(α)=[tex]\frac{component of x}{component of y}[/tex]
where α is the angle toward east of north.
Tan(α)=[tex]\frac{22.78}{57.5}[/tex]
α=21.61°