an airplane was moving with a velocity equal to 300 mph when the velocity was increased with uniform acceleration to 600 mph in 5 min. calculate the number of miles the airplane traveled during that interval of time
Vo = 300mi/h * 1600m/mi * 1h/3600s =133.3 m/s.
Vf = (600/300) * 133.33 = 266.66 m/s.
t = 5min * 60s/min = 300 s.
a = (Vf-Vo)/t = (266.66-133.33) / 300 =
0.444 m/s^2.
d = Vo*t + 0.5a*t^2,
d = 133.33*300 + 0.222*(300)^2=59,979 m
= 37.5 mi.
To calculate the distance traveled by the airplane during the interval of time, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 300 mph
Final velocity (v) = 600 mph
Time (t) = 5 min = 5/60 hours (since we need the time in hours for consistent units)
First, we need to find the acceleration (a) using the equation of motion:
v = u + a * t
Substituting the given values:
600 = 300 + a * (5/60)
Simplifying, we can solve for a:
300 = (a * 5) / 60
300 * 60 = a * 5
a = (300 * 60) / 5
a = 3600 mph^2 (acceleration in mph^2)
Now, we can substitute the values of u, a, and t into the distance formula:
distance = 300 * (5/60) + (1/2) * 3600 * (5/60)^2
Simplifying the equation:
distance = 25 + 1/2 * 3600 * (1/144)
distance = 25 + 25
distance = 50 miles
Therefore, the airplane traveled a distance of 50 miles during that interval of time.