Question

i) An aeroplan e has a run of 500 m to take
off from the runway. It starts from rest
and moves with constant acceleration to
cover the runway in 30 sec. What is the
velocity of the aeroplane at the take off ?


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Answer 1

Answer:

velocity of the plane at the take off is 33.33 m/s

Explanation:

Given:

Distance covered, s = 500 m

Time taken, t = 30 seconds

Now, from Newton's equation of motion, we have

s=ut+\frac{1}{2}at^2

where, u is the initial speed of the plane i.e 0 as starting from the rest

a is the acceleration of the plane

on substituting the values in the above equation, we have

500=0\times30+\frac{1}{2}\times\a\times30^2

or

a = 1.11 m/s²

also,

from Newton's equation of motion, we have

v = u + at

where, v is the final speed

on substituting the values in the above question, we have

v = 0 + 1.11 × 30

or

v = 33.33 m/s

Hence, the velocity of the plane at the take off is 33.33 m/s