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

V = a * t ,

101 m/s = 4.60 m/s^2 * t,
t = 101/4.6 = 22 s.

b. d = 0.5 * a * t^2 = 0.5 * 4,6 m/s^2
* (22 s)^2 = 1113 m = 1.113 km.

To solve this problem, we can use the equations of motion. The equation we will be using is:

v^2 = u^2 + 2a s

where:
v is the final velocity (0 m/s, since the plane comes to rest)
u is the initial velocity (101 m/s)
a is the acceleration (-4.60 m/s^2)
s is the distance traveled

(a) From the instant the plane touches the runway, what is the minimum time needed before it can come to rest?

Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)

Plugging in the values:
s = (0^2 - 101^2) / (2 * -4.60)

Calculating the result:
s = (-10201) / (-9.20)
s = 1111.19 m

Therefore, the minimum time needed before the plane can come to rest is approximately 1,111.19 meters.

(b) What is the actual minimum distance needed to stop the plane?

Since the plane comes to rest at the end, the distance traveled is the same as the total distance needed to stop.

Converting the distance to kilometers:
d = 1111.19 m / 1000
d = 1.11 km

Therefore, the actual minimum distance needed to stop the plane is approximately 1.11 kilometers.

To solve this problem, we can use the kinematic equation that relates final velocity (v), initial velocity (u), acceleration (a), and time (t):

v = u + at

Given:
Initial velocity (u) = 101 m/s
Acceleration (a) = -4.60 m/s² (negative because it's deceleration)

(a) To find the minimum time needed before the plane comes to rest, we need to find the time it takes for the final velocity (v) to become zero. We can rearrange the equation to solve for time (t):

0 = 101 m/s + (-4.60 m/s²) * t

Simplifying the equation:

4.60 m/s² * t = 101 m/s

t = 101 m/s / 4.60 m/s²

t ≈ 21.96 s

Therefore, the minimum time needed before the plane can come to rest is approximately 21.96 seconds. (a) The answer is 21.96 s.

(b) To find the minimum distance needed to stop the plane, we can use the equation of motion:

s = ut + (1/2)at²

Given:
Initial velocity (u) = 101 m/s
Acceleration (a) = -4.60 m/s²
Time (t) = 21.96 s (from part a)

Plugging in the values:

s = (101 m/s) * (21.96 s) + (1/2) * (-4.60 m/s²) * (21.96 s)²

s = 2222.96 m

Converting to kilometers:

s = 2222.96 m / 1000

s ≈ 2.22 km

Therefore, the actual minimum distance needed to stop the plane is approximately 2.22 kilometers. (b) The answer is 2.22 km.