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?.
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Explanation:
Initial speed of airplane u=0 m/s
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 s
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 2
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21 at
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21 at 2
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21 at 2
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21 at 2 ⇒ S=0(32.8)+0.5(3.2)(32.8)
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21 at 2 ⇒ S=0(32.8)+0.5(3.2)(32.8) 2
Initial speed of airplane u=0 m/sAcceleration of the plane a=3.2 m/s 2 and time taken t=3.28 sUsing formula, distance covered S=ut+ 21 at 2 ⇒ S=0(32.8)+0.5(3.2)(32.8) 2 =1721.3∼1720m
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