Answer:
the distance gone in the first 3 secs = 13.5 m
Step-by-step explanation:
We can find the distance gone in the first 3 secs by using the relationship between velocity and distance:
Since velocity is the rate of change in position, then the definite integral of velocity will give us the amount of distance.
[tex]\boxed{distance\ from\ a\ to\ b\ seconds=\int\limits^b_a {(velocity)} \, dx }[/tex]
Then, distance in the first 3 seconds:
[tex]\displaystyle=\int\limits^3_0 {(2t^2-t)} \, dx[/tex]
[tex]\displaystyle=\frac{2}{2+1}t^{(2+1)}-\frac{1}{1+1}t^{(1+1)}\bigg|^3_0[/tex]
[tex]\displaystyle=\frac{2}{3} t^3-\frac{1}{2} t^2\bigg|^3_0[/tex]
[tex]\displaystyle=\left[\frac{2}{3} (3)^3-\frac{1}{2} (3)^2\right]-\left[\frac{2}{3} (0)^3-\frac{1}{2} (0)^2\right][/tex]
[tex]=\bf 13.5\ m[/tex]