Answer:
60.42 m/s.
Explanation:
This problem can be solved using equation of motion
[tex]V^2 = u^2 + 2as[/tex]
where v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance moved by the body to reach final velocity v from initial velocity u.
___________________________________________________
given
u = 21m/s
a = 3.0 m/s2
s = 535 m
v, the final velocity we have to find
using the equation of motion
[tex]V^2 = u^2 + 2as\\V^2 = 21^2 + 2*3*535\\V^2 = 441 + 3210 = 3651\\\sqrt{v^2} = \sqrt{3651} \\v = 60.42[/tex]
Thus, the final velocity of airplane is 60.42 m/s.
The final velocity is 60.42 m/s
The parameters given in the question are;
initial velocity(u)= 21 m/s
acceleration(a)= 3.0 m/s²
distance(s)= 535 meters
final velocity(v)= ?
The final velocity can be calculated as follows
v²= u² +2as
v²= 21² + 2(3)(535)
v²= 441 + 3210
v²= 3651
v= √3651
v= 60.42
Hence the final velocity is 60.42 m/s
Please see the link below for more information
https://brainly.com/question/17995177?referrer=searchResults
