Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.
1 answer:
Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation
∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
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34 m/s
Explanation:
m = Mass of glider with person = 680 kg
v = Velocity of glider with person = 34 m/s
= Mass of glider without person = 680-60 kg
= Gliders speed just after the skydiver lets go
= Mass of person = 60 kg
= Velcotiy of person = 34 m/s
As the linear momentum of the system is conserved
The gliders speed just after the skydiver lets go is 34 m/s
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