Answer:
[tex]6v_0[/tex]
Explanation:
The speed of the particle increases linearly with time, so we can write it as
[tex]v(t) = kt+v_0[/tex]
where k is the a certain constant of proportionality and [tex]v_0[/tex] is the initial speed.
We also know that at t = 4 s, the speed is [tex]v=2v_0[/tex], so we can use this information to find k:
[tex]v(4) = 4k+v_0 = 2v_0 \rightarrow k = \frac{v_0}{4}[/tex]
So the complete expression of v(t) is
[tex]v(t) = \frac{v_0}{4}t+v_0 = v_0 (\frac{t}{4}+1)[/tex]
Now we integrate the quantity [tex]v(t)[/tex], and we find:
[tex]\int {v(t)} \, dt =\int (\frac{v_0}{4}t+v_0) \, dt =\frac{v_0}{8}t^2 + v_0 t[/tex]
Substituting the limit of integration t = 4, we find:
[tex]\frac{v_0}{8}4^2 + 4v_0 =2v_0+4v_0 = 6v_0[/tex]
And substituting t = 0, we find 0. So, the result of the integration from 0 to 4 is
[tex]6v_0[/tex]
