A plane with a velocity of 111. 9 meters per second at a maximum rate of 6.8 m/s^2 the plane touches the runaway what's the minimum time before it can come to rest?
A=6.8m/s^2
V=111.9m/s
t=?
To find the minimum time before the plane comes to rest, we can use the formula:
Vf = Vi + at,
where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity (Vf) will be 0 m/s since the plane has come to rest. The initial velocity (Vi) is 111.9 m/s, and the acceleration (a) is -6.8 m/s^2 (assuming the acceleration is in the opposite direction to the velocity). We need to find the time (t).
Rearranging the formula, we have:
0 = 111.9 + (-6.8)t
Now, let's solve for t:
-6.8t = -111.9
Dividing both sides by -6.8:
t = (-111.9) / (-6.8)
t ≈ 16.45 seconds
Therefore, the minimum time before the plane can come to rest is approximately 16.45 seconds.