A jet needs to reach a takeoff speed of 75.0 m/s. If the runway is 1625 meters long, what
minimum acceleration is needed? How much time will it take to reach the takeoff speed
A cannon 25m off the ground It fires horizontally at 320m/s muzzle velocity How close can it be to begin firing?
To find the minimum acceleration needed for the jet to reach its takeoff speed, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (takeoff speed) = 75.0 m/s
u = initial velocity (which we assume is 0 m/s)
a = acceleration
s = distance (runway length) = 1625 meters
Substituting the given values into the equation, we have:
(75.0 m/s)^2 = (0 m/s)^2 + 2a * 1625 meters
Simplifying the equation, we get:
5625 m^2/s^2 = 3250a
To find the minimum acceleration, isolate 'a' by dividing both sides of the equation by 3250:
a = 5625 m^2/s^2 / 3250
a ≈ 1.732 m/s^2
So, the minimum acceleration needed for the jet to reach its takeoff speed is approximately 1.732 m/s^2.
To calculate the time required to reach the takeoff speed, we can use the equation:
v = u + at
Where:
v = final velocity (takeoff speed) = 75.0 m/s
u = initial velocity (0 m/s)
a = acceleration (1.732 m/s^2, as calculated earlier)
t = time
Substituting the given values into the equation, we have:
75.0 m/s = 0 m/s + 1.732 m/s^2 * t
Simplifying the equation, we get:
75.0 m/s = 1.732 m/s^2 * t
To isolate 't', divide both sides of the equation by 1.732:
t = 75.0 m/s / 1.732 m/s^2
t ≈ 43.301 s
Therefore, it will take approximately 43.301 seconds for the jet to reach its takeoff speed.