A jet plane starts from rest with an acceleration of 3ms-2 and makes a run for 35s before taking off. What is the minimum length of the runway and what is the velocity of the jet at take off?.

Acceleration = 3 units

Time = 35s

Initial velocity = 0 units

Using s=ut+0.5a(t^2)

So,

Distance

=0.5*a*t*t (Because u=0)

=0.5*3*35*35=0.5*3*1225=1837.5 meters =1.8375 km

6

step-by-step explanation:

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.