The acceleration of a particle is constant. At t = 0, the particle is at the origin and the velocity of the particle is vo = vji
1 answer:
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 ^
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1) Initial upward acceleration:
2) Mass of burned fuel:
Explanation:
1)
There are two forces acting on the rocket at the beginning:
- The force of gravity, of magnitude , in the downward direction, where
is the rocket's mass
is the acceleration of gravity
- The thrust of the motor, T, in the upward direction, of magnitude
According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:
(1)
where a is the acceleration of the rocket.
Solving for a, we find the initial acceleration:
2)
When the rocket reaches an altitude of 5000 m, its acceleration has increased to
The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.
We can therefore rewrite eq.(1) as
where
is the new mass of the rocket
Re-arranging the equation and solving for m', we find
And since the initial mass of the rocket was
This means that the mass of fuel burned is