Answer to Question #103026 in Mechanics | Relativity for Nimra

Question

A Jet lands on a aircraft carrier at 180 mile (123 m/s) (a)what is its acceleration if it stops in 3 second (b)what is the displacement of the plane which it is stopping

Answer 1

(a) We can find the acceleration of the plane from the kinematic equation:

$v_f = v_i + at,$

here, $v_i = 123 \dfrac{m}{s}$ is the initial velocity of the plane, $v_f = 0 \dfrac{m}{s}$ is the final velocity of the plane when it stops, $a$ is the acceleration of the plane and $t$ is the time interval during which the plane is stopping.

Then, from this equation we can find the acceleration of the plane:

$a = \dfrac{v_f-v_i}{t} = \dfrac{0 \dfrac{m}{s}-123 \dfrac{m}{s}}{3s} = -41 \dfrac{m}{s^2}.$

The sign minus indicates that the jet is decelerates.

(b) We can find the displacement of the plane from another kinematic equation:

$d = \dfrac{1}{2}(v_i + v_f)t = \dfrac{1}{2} \cdot 123 \dfrac{m}{s} \cdot 3s = 184.5 m.$

Answer:

(a) $a = -41 \dfrac{m}{s^2}.$

(b) $d = 184.5 m.$