Two planes leave wichita at noon. one planes flies east 30m/h faster than the other plane, flying east at what time will they be 1200 mi apart?

After 40 hours, the planes will be 1200 mi apart.

Explanation:

Every hour the faster plane gets 30 miles more distant from the slower plane. In the first hour, the distance between the two planes will be 30 mi, in the second, 60 mi, in the third 90 mi and in the fourth hour the planes will have 120 miles distance from each other. If in four hours the distance between each plane is 120 mi, they will need ten times that amount of time to get 1200 mi (120 x10) apart. That is 40 hours.

This can also be solved using kinematic equations:

The position of the planes can be written in this way:

x = vt (slower plane)

where:

x = positon

v = velocity

t = time

We want to know at which time the planes will get a distance of 1200 mi.

Then for the faster plane the equation can be written this way:

x(faster) = v(faster)*t

The position of the faster plane will be the position of the slower plane plus 1200 mi (since the planes will be 1200 mi apart). The velocity of the faster plane is the velocity of the slower plane plus 30 mi/h. Then, we can replace x(faster) for "x + 1200 mi" (where x is the position of the slower plane) and v(faster) for "30 mi/h + v" where v is the velocity of the slower plane:

  x(faster)   =     v(faster)t

x + 1200mi = (30 mi/h + v)t

replacing x = vt (see equation for the slower plane) and distributing t.

vt + 1200 mi = 30 mi/h * t +vt

solving for t:

1200 mi = 30 mi/h * t

1200 mi / 30 mi/h = t

40 h = t

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.)