Question

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

Answer 1

Answer:

simply, find the distance by (s=u+1/2 gt²) and then distance = 1837.5m then find velocity by (v=u+gt)

Explanation:g = 3m/s²

Answer 2

The minimum length of the runway is 587.5 m, and the velocity of the jet at takeoff is 105 m/s.

The minimum length of the runway can be calculated using the equation of motion:

$d = (v_f^2 - v_i^2) / (2a)$

where d is the minimum length of the runway, v_f is the final velocity (at takeoff), v_i is the initial velocity (0 m/s since the plane starts from rest), and a is the acceleration $(3 m/s^2).$

Substituting the given values, we get:

$d = (v_f^2) / (2 * 3)$

$d = (v_f^2) / 6$

Since the plane runs for 35 seconds before taking off, we can use the equation:

$v_f = v_i + a * t$

to find the final velocity at takeoff:

$v_f = 0 + 3 * 35$

v_f = 105 m/s

Substituting this value into the equation for d, we get:

$d = (105^2) / 6$

$d = 3525 / 6$

d = 587.5 m

So, the minimum length of the runway is 587.5 m, and the velocity of the jet at takeoff is 105 m/s.

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