Physics, asked by Girijaaaa, 1 year ago

An aeroplane starts from rest with an acceleration of 3ms-2 and takes a run for 35 s before taking off. What is minimum length of the runway and with what velocity the plane took off?

Answers

Answered by siddhartharao77
178
s = ut+1/2at^2 
     
         s = 0 + 1/2 * 3 * 35 * 35

         s = 1837.5m

v = u + at

   = 0 + 3 * 35

  = 105 m/s.


Hope this helps!

Girijaaaa: Thanks
Answered by muscardinus
40

Given that,

Initial speed of an aeroplane, u = 0

Acceleration of the aeroplane, a=3\ m/s^2

Time, t = 35 s

To find,

The minimum length of the runway and with what velocity the plane took off.

Solution,

The minimum length of the runway means we need to find the distance covered. So,

d=ut+\dfrac{1}{2}at^2

u = 0

d=\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 3\times 35^2\\\\d=1837.5\ m

Let v is the took off velocity. So,

v=u+at\\\\v=3\times 35\\\\v=105\ m/s

Hence, this is the required solution.

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