Answer :
Answer:
The value of magnitude of the velocity vector of the truck relative to the SUV [tex]V_{xy}= 25.62 \frac{m}{sec}[/tex]
Explanation:
Velocity of the truck [tex]V_{x}[/tex] = 16 [tex]\frac{m}{s}[/tex]
Velocity of the SUV [tex]V_{y}[/tex] = - 20 [tex]\frac{m}{s}[/tex]
The magnitude of the velocity vector of the truck relative to the SUV is
[tex]V_{xy} =\sqrt{ V_{x} ^{2} + V_{y}^{2}}[/tex]
Put the value of [tex]V_{x}[/tex] & [tex]V_{y}[/tex] in the above equation we get
[tex]V_{xy} = \sqrt{(16)^{2} + (-20)^{2}}[/tex]
[tex]V_{xy} = \sqrt{656}[/tex]
[tex]V_{xy}= 25.62 \frac{m}{sec}[/tex]
This is the value of magnitude of the velocity vector of the truck relative to the SUV.
