A lorry of mass 4000 kg is travelling at a speed of 4 m/s.
1 answer:
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Answer:
Mininmum acceleration required =
Explanation:
Given,
Lift off speed = 125km/hr.
We know km/hr conversion factor to m/s is ,
Therefore,
Lift off speed required = m/s
Lift off speed required = 34.72222 m/s
Length of Takeoff Run given = 220 m.
So,
the flight needs to achieve its takeoff speed at a constant acceleration before it reaches the end of the 220 m runway.
Consider the formula from kinematics,
,
where,
v - final velocity of particle,
u - initial velocity of particle
a - the constant acceleration at which the particle is moving with
s - distance travelled by particle
So at minimum acceleration,it reaches the takeoff speed at the end of the runway.
Therefore in the formula,
we use, v = Lift off speed = 34.72222 m/s;
u = 0 m/s (plane starts from rest);
a = minimum acceleration of the plane
s = 220 m;
Substituting these values in the formula, we get,
a =
a = .
Therefore,the minimum acceleration of the plane required = .