a) Average velocity: 2.8 m/s
b) Average velocity: 5.2 m/s
c) Average velocity: 7.6 m/s
Explanation:
a)
The position of the car as a function of time t is given by
[tex]x(t)=\alpha t^2 - \beta t^3[/tex]
where
[tex]\alpha = 1.50 m/s^2[/tex]
[tex]\beta = 0.05 m/s^3[/tex]
The average velocity is given by the ratio between the displacement and the time taken:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
The position at t = 0 is:
[tex]x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0[/tex]
The position at t = 2.00 s is:
[tex]x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m[/tex]
So the displacement is
[tex]\Delta x = x(2)-x(0)=5.6-0=5.6 m[/tex]
The time interval is
[tex]\Delta t = 2.0 s - 0 s = 2.0 s[/tex]
And so, the average velocity in this interval is
[tex]v=\frac{5.6 m}{2.0 s}=2.8 m/s[/tex]
b)
The position at t = 0 is:
[tex]x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0[/tex]
While the position at t = 4.00 s is:
[tex]x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m[/tex]
So the displacement is
[tex]\Delta x = x(4)-x(0)=20.8-0=20.8 m[/tex]
The time interval is
[tex]\Delta t = 4.0 - 0 = 4.0 s[/tex]
So the average velocity here is
[tex]v=\frac{20.8}{4.0}=5.2 m/s[/tex]
c)
The position at t = 2 s is:
[tex]x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m[/tex]
While the position at t = 4 s is:
[tex]x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m[/tex]
So the displacement is
[tex]\Delta x = 20.8 - 5.6 = 15.2 m[/tex]
While the time interval is
[tex]\Delta t = 4.0 - 2.0 = 2.0 s[/tex]
So the average velocity is
[tex]v=\frac{15.2}{2.0}=7.6 m/s[/tex]
Learn more about average velocity:
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