A particle 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.

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