A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.
1 answer:
Answer:
The average acceleration is 16.6 m/s² ⇒ 1st answer
Explanation:
A rocket achieves a lift-off velocity of 500.0 m/s from rest in
30.0 seconds
The given is:
→ The initial velocity = 0
→ The final velocity = 500 meters per seconds
→ The time is 30 seconds
Acceleration is the rate of change of velocity of the rocket
→
where a is the acceleration, v is the final velocity, u is the initial velocity
and t is the time
→ u = 0 , v = 500 m/s , t = 30 s
Substitute these values in the rule
→ m/s²
<em>The average acceleration is 16.6 m/s²</em>
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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 ^