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
the average velocity from zero to 75.0 m/s is ... (0 + 75.0) / 2 = 37.5 m/s
time to traverse the runway is ... 1625 m / 37.5 m/s
acceleration = take off speed / traverse time
Well, to find the minimum acceleration needed, we have to remember a basic equation: acceleration = change in velocity / time. Given that the velocity needs to increase from 0 to 75.0 m/s, and the distance to cover is 1625 meters, we can calculate it:
So, the time needed to reach 75.0 m/s can be found using the formula:
time = distance / velocity
time = 1625 m / 75.0 m/s = 21.67 seconds (rounded to two decimal places)
Now, to find the minimum acceleration needed, we need to use the equation:
acceleration = change in velocity / time
acceleration = 75.0 m/s / 21.67 s = 3.46 m/s^2 (also rounded to two decimal places).
So, the minimum acceleration needed for the jet to reach that takeoff speed of 75.0 m/s is 3.46 m/s^2. Keep in mind, this is just the minimum requirement; it might need a little extra boost to account for wind resistance or perhaps to impress some flying squirrels.
To find the minimum acceleration required, you can use the following equation:
v^2 = u^2 + 2as
Where:
v = final velocity (75.0 m/s)
u = initial velocity (0 m/s, assuming the jet is starting from rest)
a = acceleration (unknown)
s = distance (1625 m)
Rearranging the equation to solve for acceleration (a), we get:
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation:
a = (75.0^2 - 0^2) / (2 * 1625)
a = 5625 / 3250
a = 1.73 m/s^2 (rounded to two decimal places)
Therefore, the minimum acceleration needed for the jet to reach a takeoff speed of 75.0 m/s is approximately 1.73 m/s^2.
To calculate the time it will take to reach the takeoff speed, you can use the following equation:
v = u + at
Where:
v = final velocity (75.0 m/s)
u = initial velocity (0 m/s)
a = acceleration (1.73 m/s^2, as calculated above)
t = time (unknown)
Rearranging the equation to solve for time (t), we get:
t = (v - u) / a
Substituting the given values into the equation:
t = (75.0 - 0) / 1.73
t = 43.35 seconds (rounded to two decimal places)
Therefore, it will take approximately 43.35 seconds for the jet to reach a takeoff speed of 75.0 m/s.
To find the minimum acceleration needed for the jet to reach the takeoff speed of 75.0 m/s, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (takeoff speed) = 75.0 m/s
u = initial velocity (0 m/s, as the jet starts from rest)
a = acceleration
s = distance (runway length) = 1625 meters
Substituting the given values into the equation, we have:
75.0^2 = 0^2 + 2a(1625)
5625 = 3250a
Now, we can solve for acceleration (a):
a = 5625 / 3250
a = 1.73 m/s^2 (rounded to two decimal places)
Therefore, the minimum acceleration required for the jet to reach a takeoff speed of 75.0 m/s is 1.73 m/s^2.
To calculate the time it takes to reach the takeoff speed, we can use the following kinematic equation:
v = u + at
Where:
v = final velocity (takeoff speed) = 75.0 m/s
u = initial velocity (0 m/s)
a = acceleration (1.73 m/s^2, as calculated earlier)
t = time
Substituting the given values into the equation, we have:
75.0 = 0 + 1.73t
Simplifying, we find:
t = 75.0 / 1.73
t ≈ 43.35 seconds (rounded to two decimal places)
Therefore, it will take approximately 43.35 seconds for the jet to reach the takeoff speed.