Aparticle moves along the x-axis with position function s(t) = ecos(x). how many times in the interval [0, 2π] is the velocity equal to 0?

It goes to zero three times

Step-by-step explanation:

s(t) = e^ cos(x)

To find the velocity, we have to take the derivative of the position

ds/dt =  -sin x e^ cos x dx/dt

Now we need to find when this is equal to 0

0 =  -sin x e^ cos x

Using the zero product property

-sin x=0   e^cos x= 0

sin x = 0

Taking the arcsin of each side

arcsin  sinx= arcsin 0

x = 0 ,pi, 2 pi

e^cos x= 0

Never goes to zero

The velocity is equal to 0 for 3 times.

Step-by-step explanation:

Given position function s = ecos(x)

Its velocity function, s' = ds/dt = e(-sinx)dx/dt

Between [0,2π], s'=0, -e(sinx)dx/dt=0

sinx=0

x=0, π, 2π

The velocity is equal to 0 for 3 times.

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

1) 2/5

2) zero=0/1

3) zero=0/2

4) -4/5

5) undefined=3/0