A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time interval and gives the cart a final speed. To reach the same speed using a force that is half as big, the force must be exerted for a time interval that is

Let F be the magnitude of the force applied to the cart, m the mass of the cart, and a the acceleration it undergoes. After time t, the cart accelerates from rest v₀ = 0 to a final velocity v. By Newton's second law, the first push applies an acceleration of

F = m a   →   a = F / m

so that the cart's final speed is

v = v₀ + a t

v = (F / m) t

If we force is halved, so is the accleration:

a = F / m   →   a/2 = F / (2m)

So, in order to get the cart up to the same speed v as before, you need to double the time interval t to 2t, since that would give

(F / (2m)) (2t) = (F / m) t = v

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.