An airplane, traveling 142.5 km/hr at the moment it touches the runway, runs off the end of the runway still traveling at 36.7 km/hr. If the plane’s rate of deceleration was -1.6 m/s2, how long and how far did the plane travel along the runway?
To find the time and distance traveled by the plane along the runway, we can use the equations of motion.
Let's first convert the given velocities from km/hr to m/s:
Initial velocity (u) = 142.5 km/hr
Final velocity (v) = 36.7 km/hr
Acceleration (a) = -1.6 m/s^2
Converting km/hr to m/s:
1 km/hr = 1000 m/3600 s
Therefore,
u = 142.5 km/hr * (1000 m/3600 s) = 39.58 m/s
v = 36.7 km/hr * (1000 m/3600 s) = 10.19 m/s
Now, using the equation of motion:
v^2 = u^2 + 2as
We need to find "s" (distance) and "t" (time).
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values, we get:
s = (10.19^2 - 39.58^2) / (2 * -1.6)
Calculating the distance (s):
s = (-329.8361) / (-3.2)
s = 103.01 meters (rounded to two decimal places)
To find the time (t), we can use the equation:
v = u + at
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (10.19 - 39.58) / (-1.6)
Calculating the time (t):
t = (-29.39) / (-1.6)
t = 18.37 seconds (rounded to two decimal places)
Therefore, the plane traveled a distance of approximately 103.01 meters along the runway and the time taken was approximately 18.37 seconds.
idek help me.
1 km/hr = 1000/3600 = 5/18 m/s
time taken to decelerate:
t = (142.5-36.7)(5/18)/1.6 = 18.368 s
In that time, it traveled
s = 142.5t - 0.8t^2 meters