A3.2-kg object is subjected to two constant forces. f_1 = (1.9 n)i + (-1.9 n)j and f_2= (3.8 n)i +(-10.1 n)j. the object is at rest at origin at lime t = 0.
a) what is the object's acceleration?
b) what is its velocity at time t = 3.0 s?
c) where is the object at time t = 3.0 s?

A)

a= 4.15 m/s² : Magnitude of the acceleration

β= 64.59° : Direction of the acceleration with respect to the horizontal

B)

v = 12.45 m/s : Magnitude of the velocity

β= 64.59°: Direction of the velocity with respect to the horizontal

C)

d = 18.675 m : Magnitude of the displacement

β= 64.59° : Direction of the displacement with respect to the horizontal

Explanation:

A)Data

m=3.2 kg

F₁= (1.9 N)i + (-1.9 N)j

F₂= (3.8 N)i +(-10.1 N)j.

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

We apply the (Formula 1)

F₁ + F₂= m*a

( (1.9 N)i + (-1.9 N)j)+((3.8 N)i +(-10.1 N)j) =  3.2* a

5.7 i - 12 j =  3.2* a

a=( 5.7/ 3.2) i - (12/3.2) j

a= 4.15 m/s² : Magnitude of the acceleration

β= 64.59°: Direction of the acceleration with respect to the horizontal

B)-C) Because the object moves with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

a= 4.15 m/s²

t = 0,

t = 3.0 s

B) we apply the formula (2)

vf= v₀+at  

vf= 0+(4.15)(3)

vf= 12.45 m/s  :magnitude of the velocity

β= 64.59°: Direction of the velocity with respect to the horizontal

C) we apply the formula (3)

d= v₀t+ (1/2)*a*t²

d= 0+ (1/2)(4.15)(3)²

d=18.675 m :Magnitude of the displacement

β= 64.59°: Direction of the displacement with respect to the horizontal

the answer is 46 mls because when you add the speed per second it increses the speed

the answer would be   dignified - apex