Answer:
Distance, d = 1225 meters
Explanation:
It is given that,
Initial speed of the jet, u = 70 m/s
Acceleration of the jet, [tex]a=-2\ m/s^2[/tex]
It can be assumed to find the distance traveled on the runway as it is brought to rest. Let the distance is given by d. Also, the final speed, v = 0
d can be calculated using the third equation of kinematics. The formula is given by :
[tex]d=\dfrac{v^2-u^2}{2a}[/tex]
Substituting the values in above formula.
[tex]d=\dfrac{0-(70)^2}{2\times (-2)}[/tex]
d = 1225 meters
So, the distance covered by the jet as it brought to rest is 1225 meters. Hence, this is the required solution.
