A jet plane lands with a speed of 87 m/s and can accelerate at a maximum rate of -5.30 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? What is the actual minimum distance needed to stop the plane?

a. t = (Vf - Vo) / a,

t = (0 -87) / -5.30 = 16.4s.

b. d = (Vf^2 - Vo^2) / 2a,
d = (0 - (87)^2) / -10.6 = 714.1s.

To find the minimum time needed before the plane comes to rest, we can use the equation of motion:

v = u + at

Where:
v = final velocity (0 m/s, since the plane comes to rest)
u = initial velocity (87 m/s)
a = acceleration (-5.30 m/s^2)
t = time

Solving for t:

0 = 87 + (-5.30)t

-5.30t = -87

t = -87 / -5.30

t ā‰ˆ 16.42 seconds

So the minimum time needed before the plane can come to rest is approximately 16.42 seconds.

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

s = ut + 0.5at^2

Where:
s = distance
u = initial velocity (87 m/s)
t = time (16.42 seconds)
a = acceleration (-5.30 m/s^2)

Solving for s:

s = 87(16.42) + 0.5(-5.30)(16.42)^2

s ā‰ˆ 1426 meters

Therefore, the actual minimum distance needed to stop the plane is approximately 1426 meters.

To find the minimum time needed for the jet plane to come to rest, we can use the formula:

š‘£ = š‘¢ + š‘Žš‘”

where š‘£ is the final velocity (which is 0 m/s when the plane comes to rest), š‘¢ is the initial velocity (87 m/s), š‘Ž is the acceleration, and š‘” is the time.

In this case, the acceleration is given as -5.30 m/sĀ² (negative sign indicates deceleration).

Substituting the given values into the formula, we have:

0 = 87 - 5.30š‘”

Solving for š‘”:

5.30š‘” = 87

š‘” = 87 / 5.30

š‘” ā‰ˆ 16.42 seconds

Therefore, the minimum time needed for the jet plane to come to rest is approximately 16.42 seconds.

To find the minimum distance needed to stop the plane, we can use the formula:

š‘  = š‘¢š‘” + 0.5š‘Žš‘”Ā²

where š‘  is the distance traveled, š‘¢ is the initial velocity (87 m/s), š‘Ž is the acceleration (-5.30 m/sĀ²), and š‘” is the time.

Substituting the known values:

š‘  = 87(16.42) + 0.5(-5.30)(16.42)Ā²

š‘  ā‰ˆ 714.27 meters

Therefore, the minimum distance needed to stop the plane is approximately 714.27 meters.