A space aircraft is launched straight up.The aircraft motor provides a constant acceleration for 10 seconds,then the motor stops.The aircraft's altitude 15 seconds after launch is 2 km.ignore air friction .what is the acceleration,maximum speed reached in km/h and the speed(in km/h)of the aircraft as it passes through a cloud 4 km above the ground.
A vertical distance covered for t₁=10 s of accelerated motion
is h₁=at₁²/2. The speed at this height is v₁=at₁.
The distance covered at decelerated motion during t₂=15- t₁=15-10=5 s
is h₂=v₁t₂-gt₂²/2.
H= 2000 m
H=h₁+h₂=at₁²/2 + v₁t₂ - gt₂²/2=
=at₁²/2 + at₁t₂ - gt₂²/2.
a(t₁² +2 t₁t₂) =2H+ gt₂²
If g=10 m/s²,
a={2H+ gt₂²}/(t₁² +2 t₁t₂)=
={4000+10•25}/(100+2•10•5)=21.25 m/s²
Max velocity is v₁=at₁=21.25•10=212.5 m/s=765 km/h
The aircraft covered the distance h₁=at₁²/2 =21.25•10²/2 =1062.5 m at accelerated motion and the distance h₃=4000- h₁=4000-1062.5 =2937.5 m at decelerated motion. This motion takes the time t₃.
h₃=v₁t₃-gt₃²/2.
gt₃² -2 v₁t₃ +2h₃=0
Solve for t₃
v₃= v₁-gt₃=…
But for this given data, I’ve calculated the height h₀ at which the velocity becomes zero
0=v₁ -gt => t=v₁/g=212.5/10=21.25 s
h₀=v₁t-gt²/2=212.5•21.25 - 10•21.25²/2 = 2257.8 m. =>
The max height is h₁+h₀=1062.5+2257.8 =3320.3 m => Aircraft con’t reach the height 4000 m. Anyhow, try to check me on my math…
To find the answers, we can use the kinematic equations of motion.
1. To find the acceleration, we can use the equation:
Δy = V0t + (1/2)at^2
Here, Δy is the change in altitude, V0 is the initial velocity, t is the time, and a is the acceleration.
In this case, the change in altitude is 2 km and the time is 15 seconds. Since the motor accelerates the aircraft for the first 10 seconds, we can consider the time after the motor stops as 15 - 10 = 5 seconds.
Plugging the values into the equation, we get:
2 km = V0 * 15 s + (1/2) * a * (15 s)^2
Now, we need to solve for acceleration (a).
Rearranging the equation, we get:
a = [2 km - V0 * 15 s] / [(1/2) * (15 s)^2]
2. To find the maximum speed reached, we need to calculate the velocity at the end of the 10-second acceleration period.
We can use the equation:
V = V0 + at
Here, V is the final velocity.
The initial velocity, V0, is 0 km/h (since the aircraft starts from rest).
The acceleration, a, is calculated from the previous step.
The time, t, is 10 seconds (since the motor provides constant acceleration for 10 seconds).
Plugging the values into the equation, we get:
V = 0 km/h + a * 10 s
3. To find the speed (in km/h) of the aircraft as it passes through a cloud 4 km above the ground, we can use the equation of motion:
V^2 = V0^2 + 2aΔy
Here, V is the final velocity, V0 is the initial velocity, a is the acceleration, and Δy is the change in altitude.
The change in altitude, Δy, is 4 km (since the cloud is 4 km above the ground).
The acceleration, a, is calculated from the first step.
Plugging the values into the equation, we get:
V^2 = 0 + 2a * 4 km
To calculate the values, we need the initial velocity (V0). It is not provided in the given question. Please provide the initial velocity of the aircraft.