What is the initial velocity of a runner who reaches a velocity of 2.41 m/s due east in 2.50 s, 2 assuming his average acceleration during that period of time is 0.130 m/s , east?

v= 2.74m/s

Explanation:

We can solve for the initial velocity using the first equation of motion

where

v= final velocity

u= initial velocity

a= acceleration

t= time in seconds

Given data

final velocity= 2.41 m/s

initial velocity v =?

time t= 2.5 seconds

acceleration a=

substituting we have

v= 2.41+0.13(2.5)

v= 2.41+0.325

v= 2.735m/s

v= 2.74m/s

the final velocity is  2.74 m/s

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge