An airplane taking off from a landing field has run of 400 m. If it starts from rest and accelerates its motion uniformly and makes the run in 20 seconds. with what velocity did it take off
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To find the velocity at which the airplane took off, we can use the equation of motion that relates uniform acceleration (a), initial velocity (u), time (t), and displacement (s):
s = ut + (1/2)at^2
In this case, the initial velocity (u) is 0 m/s (since it starts from rest), the time (t) is 20 seconds, and the displacement (s) is 400 meters. We need to solve for the acceleration (a).
Rearranging the equation, we get:
s = (1/2)at^2
400 = (1/2)a(20^2)
400 = 200a
Divide both sides by 200:
a = 400/200
a = 2 m/s^2
Now that we have the acceleration, we can find the final velocity (v) using the equation:
v = u + at
Since the initial velocity (u) is 0, the equation simplifies to:
v = at
v = 2 * 20
v = 40 m/s
Therefore, the airplane took off with a velocity of 40 m/s.
To find the velocity at which the airplane took off, we need to use the kinematic equation that relates velocity, distance, acceleration, and time. The equation we will use in this case is:
v = u + a*t
Where:
v = final velocity
u = initial velocity (which is 0 in this case, as the airplane starts from rest)
a = acceleration
t = time
We are given:
Distance (d) = 400 m
Time (t) = 20 s
Initial velocity (u) = 0 m/s
We need to find the final velocity (v).
First, we can calculate the acceleration (a) using the formula:
a = (2*d) / (t^2)
Substituting the given values:
a = (2 * 400 m) / (20 s)^2
a = 800 m / 400 s^2
a = 2 m/s^2
Now we can substitute the values of u, a, and t into the equation:
v = u + a*t
v = 0 m/s + (2 m/s^2) * 20 s
v = 0 m/s + 40 m/s
v = 40 m/s
Therefore, the airplane took off with a velocity of 40 m/s.
average speed = 400/20 = 20 m/s^2
constant acceleration (v = a t + starting speed) so begin speed is 0 and end speed is 40 m/s