suppose an airplane, landing on a runway with an initial speed of 45 m/s, slows down at an acceleration rate of -5m/s/s. What will the speed of the airplaine be one second after landing?
V = Vo + at = 45 + (-5)*1 = 40m/s.
To find the speed of the airplane one second after landing, we can use the equation:
v = u + a*t
Where:
v = final velocity (speed of the airplane after 1 second)
u = initial velocity (speed of the airplane before landing)
a = acceleration
t = time (in this case, 1 second)
Given:
u = 45 m/s
a = -5 m/s^2
t = 1 second
Plugging in these values into the equation, we have:
v = 45 m/s + (-5 m/s^2) * 1 second
Simplifying the equation, we have:
v = 45 m/s - 5 m/s
v = 40 m/s
Therefore, the speed of the airplane one second after landing will be 40 m/s.
To find the speed of the airplane one second after landing, we can use the equation of motion:
vf = vi + at
where:
vf = final velocity (speed of the airplane after one second)
vi = initial velocity (speed of the airplane before landing)
a = acceleration
t = time
Given:
vi = 45 m/s (initial speed of the airplane)
a = -5 m/s² (acceleration rate)
t = 1 second (time)
Now we substitute the given values into the equation:
vf = 45 m/s + (-5 m/s² * 1 s)
First, we calculate -5 m/s² * 1 s, which gives us -5 m/s.
Then, we substitute that back into the equation:
vf = 45 m/s - 5 m/s
Finally, we perform the subtraction:
vf = 40 m/s
Therefore, the speed of the airplane one second after landing will be 40 m/s.