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.
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?
1 answer
1 answer