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

Explanation:

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

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

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

So the complete expression of v(t) is

Now we integrate the quantity , and we find:

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

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

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

not too sure thinking all accelerations are pointed down