Question

A particle moves in a straight line. Its speed(v)

Answer 1

Answer:

$6v_0$

Explanation:

The speed of the particle increases linearly with time, so we can write it as

$v(t) = kt+v_0$

where k is the a certain constant of proportionality and $v_0$ is the initial speed.

We also know that at t = 4 s, the speed is $v=2v_0$, so we can use this information to find k:

$v(4) = 4k+v_0 = 2v_0 \rightarrow k = \frac{v_0}{4}$

So the complete expression of v(t) is

$v(t) = \frac{v_0}{4}t+v_0 = v_0 (\frac{t}{4}+1)$

Now we integrate the quantity $v(t)$, and we find:

$\int {v(t)} \, dt =\int (\frac{v_0}{4}t+v_0) \, dt =\frac{v_0}{8}t^2 + v_0 t$

Substituting the limit of integration t = 4, we find:

$\frac{v_0}{8}4^2 + 4v_0 =2v_0+4v_0 = 6v_0$

And substituting t = 0, we find 0. So, the result of the integration from 0 to 4 is

$6v_0$