A jet plane lands with a speed of 101 m/s and can accelerate at a maximum rate of -4.60 m/s2 as it comes to rest.
(a) From the instant the plane touches the runway, what is the minimum time needed before it can come to rest?
___s
b)What is the actual minimum distance needed to stop the plane?
___km
To find the minimum time needed for the jet plane to come to rest, we need to determine the time it takes for the plane to decelerate from its initial speed to zero.
(a) To find the time, we can use the equation of motion:
v = u + at
Where:
v = final velocity (zero in this case)
u = initial velocity (101 m/s)
a = acceleration (-4.60 m/s^2)
t = time
Rearranging the equation to solve for time:
t = (v - u) / a
Substituting the values:
t = (0 - 101) / -4.60
Calculating:
t ≈ 21.957 s
Therefore, the minimum time needed before the jet plane can come to rest is approximately 21.957 seconds.
(b) To find the minimum distance needed to stop the plane, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (101 m/s)
t = time (21.957 s)
a = acceleration (-4.60 m/s^2)
Substituting the values:
s = (101 × 21.957) + (1/2 × -4.60 × (21.957)^2)
Calculating:
s ≈ -2344.388 m
Since distance cannot be negative in this context, we take the absolute value:
s ≈ 2344.388 m
Converting the distance to kilometers:
s ≈ 2.344 km
Therefore, the actual minimum distance needed to stop the plane is approximately 2.344 kilometers.