Graph B represents the velocity of the sphere changes over time when falling with constant acceleration.
- Acceleration is the measure of how quickly a body's velocity varies with regard to time, and constant acceleration occurs when a body's velocity changes proportionately over a period of time, or at a constant rate. It measures in m/s2.
- It is claimed that a body has continual positive acceleration when it begins to move with an initial velocity of zero and gradually increases to a positive value over time.
- Constant positive acceleration is demonstrated by a ball falling freely in a vertical direction.
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It is c that’s what it is that’s the answer
The rope will remain taut until the particle makes 79⁰ angle.
<h3>Change in kinetic energy of the particle</h3>
The change in kinetic energy of the particle is calculated as follows;
ΔK.E = K.Ei - K.Ef
Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%
ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E
- let the vertical displacement of the particle = h
- horizontal length = side of the prism = a
- hypotenuse side = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>
K.E = P.E
0.7K.E = mgh
0.7(¹/₂mv²) = mg(Lsinθ)
0.7(v²) = 2g(Lsinθ)
from third kinematic equation;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2g(a tanθ)
0.7(2g(a tanθ)) = 2g(Lsinθ)
0.7(a tanθ) = Lsinθ
0.7a/L = sinθ/tanθ
0.7a/L = cosθ
(0.7 x 0.8)/(3) = cosθ
0.1867 = cosθ
θ = cos⁻¹(0.1867)
θ = 79⁰
Thus, the rope will remain taut until the particle makes 79⁰ angle.
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Answer:
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 ^
Answer:
Acceleration: -9.8 m/s^2
Velocity: -28.2 m/s
Displacement: 143.1 m
Explanation:
The acceleration of gravity for any object close to earth is approximately -9.8 m/s^2.
Now, to find the velocity after 9 seconds, we can use a kinematics formula, where x is the final velocity:
<em>Final Velocity = Initial Velocity + Acceleration * Time</em>
x = 60 + -9.8*9
x = 60 - 88.2
x = -28.2
The velocity is -28.2 m/s.
Lastly, to find the displacement, we can use another kinematics formula, where y is the displacement:
<em>Displacement = (Final Velocity + Initial Velocity)/2 * Time</em>
y = (-28.2 + 60)/2 * 9
y = 143.1
The displacement is 143.1 meters.
