At the end of 3N force acts on the object during time interval of 1.5seconds with force acting towards right. A constant force of 4N to left is applied for 3seconds. What is the velocity at the end of the 3seconds ?

v_f = -7.5 m / s

Explanation:

Let's analyze this exercise a little, two forces that act on a body for different time intervals are indicated, each force creates an impulse and since this is a vector quantity we must add in the form of vectors. The net momentum is

we assume that the direction to the right is positive

             I = I₁ + I₂

             I = F₁ Δt₁ - F₂ Δt₂

             I = 3  1.5 - 4  3

             I = -7.5   N s

now let's use the relationship between momentum and momentum, suppose the object starts from rest (vo = 0)

             I = Δp

             I = m (v_f - v₀)

             v_f = I / m

              v_f = -7.5 / m

to finish the calculation we must assume a mass m = 1 kg

              v_f = -7.5 m / s

the negative sign in the body is moving to the left

be subject to a force andchange its motion

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.