Just after hitting the runway, a plane has a speed of 200.0 km/h. The plane decelerates to a stop at a constant rate.

a. What is its average speed on the runway in m/s?
b. If the plane takes 21 seconds to come to a stop, how much of the runway
does it cover during its deceleration?

Vo = 200km/h = 200,000m/3600s=55.56 m/s

a. Vavg. = (V+Vo)/2 = (0+55.56)/2 = 27.8 m/s.

b. V = Vo + a*t = 0
55.56 + a*21 = 0
21a = -55.56
a = -2.65 m/s^2

b.V^2 = Vo^2 + 2a*d
d = (V^2-Vo^2)/2a = (0-(55.56^2))/-5.3 =
582 m.

To find the answers, we need to convert the given values to the appropriate units and then use the relevant formulas. Let's solve each part step by step:

a. To find the average speed on the runway in m/s, we need to convert the initial speed from km/h to m/s. We can do this by multiplying the value by 1000/3600 (since there are 1000 meters in a kilometer and 3600 seconds in an hour).

Given initial speed = 200.0 km/h
Converting to m/s: 200.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 55.56 m/s (rounded to two decimal places)

Therefore, the average speed of the plane on the runway is 55.56 m/s.

b. To find how much of the runway the plane covers during its deceleration, we need to find the distance traveled. We can use the equation:

distance = (initial speed * time) + (0.5 * acceleration * time^2)

Given:
Initial speed = 200.0 km/h (converted to m/s = 55.56 m/s, as calculated in part a)
Time = 21 seconds

Since the plane is coming to a stop, its final speed is zero (0 m/s), and its acceleration is negative (deceleration). We can rearrange the equation to solve for distance:

distance = (0 m/s * 21 s) + (0.5 * acceleration * (21 s)^2)

Since the plane is decelerating and coming to a stop, we can assume its final acceleration is constant. Let's solve for acceleration first:

final speed = initial speed + (acceleration * time)
0 m/s = 55.56 m/s + (acceleration * 21 s)
acceleration * 21 s = -55.56 m/s
acceleration = -55.56 m/s / 21 s = -2.65 m/s^2 (rounded to two decimal places)

Now, we can substitute the values in the distance equation:

distance = (0 m/s * 21 s) + (0.5 * -2.65 m/s^2 * (21 s)^2)
distance = 0 m + (0.5 * -2.65 m/s^2 * 441 s^2)
distance = 0 m - (1.32 m/s^2 * 441 s^2)
distance = -1.32 m/s^2 * 441 s^2

Since distance cannot be negative in this context, we take the magnitude by simply removing the negative sign:

distance = 1.32 m/s^2 * 441 s^2 = 581.52 m^2 (rounded to two decimal places)

Therefore, the plane covers approximately 581.52 meters during its deceleration.