In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the
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Answer:
The distance before stopping is 1.52 m,
velocity is 4.0 m/s on the y-axis
Explanation:
The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.
a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
When the particle stops, we calculate its 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 returning.
b) The velocity comprises both x and y components.
For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
Thus, the velocity reads as
V = (0 x^ + 4.0 y^) m/s
c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for 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 construct two graphs for position against time, one for the x-axis and another for the y-axis
Chart for graphing
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
