At the end of a race a runner decelerates from a velocity of 7.00 m/s at a rate of 0.300 m/s2.

x=vt+1/2at^2

x=7(15)-1/2(4)(15^2)
x=71.25

Two fixed charges 1.07mC and -3.28mC are 61.8 cm apart. Where the third charge may be placed so that no net force acts on it?

To find the distance covered by the runner during deceleration, you can use the equation of motion:

v^2 = u^2 + 2as

Where:
- v is the final velocity (0 m/s, since the runner comes to a stop)
- u is the initial velocity (7.00 m/s)
- a is the acceleration/deceleration (-0.300 m/s^2, since it is deceleration)
- s is the distance covered

Rearranging the formula to solve for the distance (s):

s = (v^2 - u^2) / (2a)

Plugging in the given values:

s = (0^2 - 7.00^2) / (2 * -0.300)

Simplifying:

s = (-49.00) / (-0.600)

s = 81.66 m

Therefore, the runner covers a distance of 81.66 meters during deceleration at a rate of 0.300 m/s^2.