Question

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

increases linearly with time. If the initial speed of
the particle is vo and its speed at t = 4 s is 2V
then integration of vdt from 0 to 4 is
(1) Zvo
(3) 6v
(2) 8vo
(4) AVO

Answer 1

`6v_0`

Step-by-step 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 = (v_0)/(4)`

So the complete expression of v(t) is

`v(t) = (v_0)/(4)t+v_0 = v_0 ((t)/(4)+1)`

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

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

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

`(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`