Aparticle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗ =(−1.90 m/s2)xˆ+(3.20 m/s2)yˆa→=(−1.90 m/s2)x^+(3.20 m/s2)y^ . (a) how far does the particle move in the x direction before turning around? (b) what is the particle’s velocity at this time? (c) plot the particle’s position at t=0.500 st=0.500 s , 1.00 s, 1.50 s, and 2.00 s. use these results to sketch position versus time for the particle.

distance stop 1.52m,

velocity  4.0 m/s y^

Explanation:

The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.

a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)

     Vfₓ = V₀ₓ + aₓ t  

     t = -  V₀ₓ/aₓ

     t = - 2.4/(-1.9)

     t=  1.26 s

At this time the particle stops, let's find his position

     X1 = V₀ₓ t + ½ aₓ t²

     X1= 2.4 1.26 + ½ (-1.9) 1.26²

     X1= 1.52 m

At this point the particle begins its return

b) The velocity has component x and y

   As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis

    Vfy= Voy + ay t

    Vfy= 0 + 3.2 1.26

    Vfy = 4.0 m/s

the velocity is  

    V = (0 x^ + 4.0 y^) m/s

c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations

      X = V₀ₓ t+ ½  aₓ t²

      Y = Voy t + ½  ay t²

      X= 2.4 t + ½ (-1.9) t²

      Y= 0 + ½ 3.2 t²

      X= 2.4 t – 0.95 t²

      Y=   1.6 t²

With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis

                       Chart to graph

              Time (s)     x(m)            y(m)

                 0                0               0

                 0.5             0.960       0.4

          1       1.45          1.6

                 1.50      1.46      3.6

                2.00      1.00      6.4