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.
1 answer:
Answer:
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
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