The acceleration of a particle is constant. At t = 0, the particle is at the origin and the velocity of the particle is vo = vji+ v2j. At time t = T, the velocity of the particle is v = v3j. Here v1, v2, and v3 are constants with dimensions of length divided by time. All answers should be written in terms of vj, v2, V3, T, and the unit vectors i and j. Part (a) What is the particle's acceleration vector?
Part (b) What is the particle's position vector at t= 2T?
Part (c) What is the particle's velocity vector at t =2T?

a) a = i ^ + j^, b) r = 2 v₃ T j ^, c)    v = -v₁ i ^ + (2 v₃ - v₂) j ^

Explanation:

This is a two-dimensional kinematics problem

a) Let's find the acceleration of the body, for this let's use a Cartesian coordinate system

X axis

initial velocity v₀ₓ = v₁ for t = 0, velocity reaches vₓ = 0 for t = T, let's use

          vₓ = v₀ₓ + aₓ t

we substitute

          for t = T

           0 = v₁ + aₓ T

           aₓ = - v₁ / T

y axis  

the initial velocity is = v₂ at t = 0 s, for time t = T s the velocity is v_{y} = v₃

             v₃ = v₂ + a_{y} T

              a_{y} =

therefore the acceleration vector is

             a = i ^ + j^

b) the position vector at t = 2T, we work on each axis

X axis

             x = v₀ₓ t + ½ aₓ t²

we substitute

             x = v₁ 2T + ½ (-v₁ / T) (2T)²

              x = 2v₁ T - 2 v₁ T

              x = 0

Y axis  

             y = t + ½ a_{y} t²

             y = v₂ 2T + ½ 4T²

             y = 2 v₂ T + 2 (v₃ -v₂) T

            y = 2 v₃ T

the position vector is

            r = 2 v₃ T j ^

c) the velocity vector for t = 2T

X axis

            vₓ = v₀ₓ + aₓ t

we substitute

           vₓ = v₁ - 2T = v₁ - 2 v₁

           vₓ = -v₁

Y axis  

            = v_{oy} + a_{y} t

           v_{y} = v₂ + 2T

           v_{y} = v₂ + 2 v₃ - 2v₂

           v_{y} = 2 v₃ - v₂

the velocity vector is

           v = -v₁ i ^ + (2 v₃ - v₂) j ^

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i think the answer is a and b

i wish you luck