Answer:
13.2 m/2
Explanation:
As airplane accelerates uniformly:
final velocity^2 = initial velocity^2 + 2 * acceleration * displacement
final velocity is given as +150.0
initial velocity is given as +10.0
displacement is given as 850.0
Substituting into equation:
150^2 = 10*2 + 2 * acceleration * 850
22500 - 100 = 2 * acceleration * 850
acceleration = 22400/2/850 = 13.18 =13.2 m/s
The time it took the airplane to go that distance is 10.62 seconds.
Given the following data;
To find how long (time) it took the airplane to go that distance:
First of all, we would determine the acceleration of the airplane by using the third equation of motion;
[tex] V^2 = U^2 + 2aS [/tex]
[tex] 150^2 = 10^2 + 2a(850) [/tex]
[tex] 22500 = 100 + 1700a [/tex]
[tex] 1700a = 22500 - 100 [/tex]
[tex] 1700a = 22400 [/tex]
[tex] Acceleration, \; a = \frac {22400}{1700} [/tex]
Acceleration, a = 13.18 [tex] m/s^2 [/tex]
Next, we would solve for the time to go that distance by using the first equation of motion;
[tex] V = U + at [/tex]
[tex] 150 = 10 + 13.18t [/tex]
[tex]13.18t = 150 - 10 [/tex]
[tex]13.18t = 140 [/tex]
[tex] Time, \; t = \frac {140}{13.18} [/tex]
Time, t = 10.62 seconds.
Therefore, the time it took the airplane to go that distance is 10.62 seconds.
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