Answer:
x_total = 600 m
Explanation:
This is an exercise and kinematics, let's find the time it takes to reach the velocity 20 m / s
v = v₀ + a t
as part of rest v₀ = 0
t = v / a
t = 20/2
t = 10 s
let's find the distance traveled in this time
x₁ = vo t + ½ a t2
x₁ = 0 + ½ 2 10²
x₁ = 100 m
The remaining time is
t₂ = 35 - t
t₂ = 35 - 10
t₂ = 25 s
as in this range it has a constant speed
v = x₂ / t₂
x₂ = v t₂
x₂ = 20 25
x₂ = 500 m
the total distance traveled is
x_total = x₁ + x₂
x_total = 100 + 500
x_total = 600 m
Answer:
300 m
Explanation:
The train accelerate from the rest so u = 0 m/sec
Final speed that is v = 80 m/sec
Time t = 30 sec
The distance traveled by first plane = 1200 m
We know the equation of motion 
where s is distance a is acceleration and u is initial velocity
Using this equation for first plane 
As the acceleration is same for both the plane so a for second plane will be 2.67 
The another equation of motion is 
using this equation for second plane 
s = 300 m
Answer:
The radius of the loop is 20.66 km
Explanation:
let the radius of the loop be r
mass of airplane is m
At the top, the pilot experiences two radial forces, which are
1) Gravitational force is mg
2) Centrifugal forces mv²/r out of the center
When the pilot experiences no weight,
then, mg = mv²/r
r = v² / g
= 450² / 9.8
= 20.66 x 10³3
= 20.66 km
