Final answer:
The distance a plane flies in 13 seconds while accelerating from 144 m/s to 180 m/s is calculated using the kinematic equation for uniformly accelerated motion.
The calculated distance is 2106 meters, which suggests that there might be a mistake in the question or the options provided.
Step-by-step explanation:
To determine how far a plane flies in 13 seconds while its velocity is changing from 144 m/s to 180 m/s, we can use the formula for the distance covered under constant acceleration:
S = ut + (1/2)at2
Where:
u is the initial velocity,
a is the acceleration, and
t is the time.
First, we need to calculate the acceleration:
a = (v - u) / t
Here, v is the final velocity and t is the time. Plugging in the values:
a = (180 m/s - 144 m/s) / 13 s = 2.769 m/s2
Now, we use the first part of the equation with initial velocity (u = 144 m/s) and time (t = 13 s) to find the distance:
S = 144 m/s * 13 s + (1/2) * 2.769 m/s2 * (13 s)2
After calculating:
S = 1872 m + (1/2) * 2.769 m/s2 * 169 s2 = 1872 m + 234 m
S = 2106 m
The options provided in the question seem to be incorrect as none matches the calculated distance of 2106 m. There might have been a mistake in the question or the options given.