Answer:
The average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex] is 32.5 m/s
Explanation:
v = u +at bhjklj kj h
x = (u + v / 2 )t
Average velocity is given by
[tex]V_{avg} = \frac{Displacement}{Time} \\V_{avg} = \frac{x_{2} - x_{1} }{t_{2} - t_{1} }[/tex]
From the question,
Initial speed, u = 28 m/s
Final speed, v = 37 m/s
Time, t = 10 secs
From the formula
[tex]x = (\frac{u + v}{2})t\\[/tex]
where [tex]x[/tex] is the displacement
Put the given values into the equation to find the displacement [tex]x[/tex]
[tex]x = (\frac{28 + 37}{2}) 10\\ x = (\frac{65}{2})10\\ x = (\frac{32.5}{10})\\ x = 325 m[/tex]
Now, for the average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex]
[tex]V_{avg} = \frac{Displacement}{Time} \\V_{avg} = \frac{x_{2} - x_{1} }{t_{2} - t_{1} }[/tex]
[tex]V_{avg,x} = \frac{325 - 0}{10 - 0}\\V_{avg,x} = \frac{325}{10}\\ V_{avg,x} = 32.5 m/s[/tex]
Hence, the average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex] is 32.5 m/s
The average x component of velocity during the maneuver of the truck is 32.5 m/s.
Velocity:
The term velocity of any object is defined as the ratio of displacement covered by any object to the time taken by the object to displace.
Given data:
The magnitude of initial speed is, u = 28 m/s.
The magnitude of final speed is, v = 37 m/s.
The time interval is, t = 10 s.
The displacement covered by the truck is calculated as,
[tex]d = \dfrac{v+u}{2} \times t\\\\ d = \dfrac{37+28}{2} \times 10\\\\ d = 325 \;\rm m[/tex]
Now, the expression for the x-component of the average velocity is given as,
[tex]v_{x}=\dfrac{d}{t}[/tex]
Solving as,
[tex]v_{x}=\dfrac{325}{10}\\\\ v_{x}=32.5 \;\rm m/s[/tex]
Thus, we can conclude that the average x component of velocity during the maneuver of truck is 32.5 m/s.
Learn more about the average velocity here:
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