Salim has a paper route. In the winter time, he puts his newspapers on a toboggan to pull
FICIT rough the snow. The combined mass of the papers and the toboggan is 24 kg. The
coefficient of static friction between the toboggan and the snow is 0.18. The coefficient of kiner
friction between the toboggan and the snow is 0.10. (3 marks)
a) with how much force must Salim pull in order to get the toboggan to start moving?
b) How much force is needed to keep the toboggan moving at a constant speed?

You need to be more specific.  i really don't know what a "motion map" is.you can draw a graph of an object's position, distance, displacement, speed,velocity, or acceleration versus time, or you can draw a graph of the object's x-position versus its y-position and not give any information about time at all.if the graph presents position or displacement against time,  then constant velocity is a straight line . . i.e. a line that has  constant slope.if the graph shows speed against time, then constant velocityis a horizontal piece of line.if the graph shows acceleration against time, then constant  velocity is a piece of line that lies on the x-axis . . it's value is zero. 

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