An airplane originally at rest on a runway accelerates uniformly at 6.2 m/s2 for 12 sec. During this 12 sec-interval the air plane travels an approximate distance of?

To find the approximate distance traveled by the airplane, we can use the kinematic equation:

s = ut + (1/2)at^2

Where:
s is the distance traveled,
u is the initial velocity (0 m/s, since the airplane was originally at rest),
a is the acceleration (6.2 m/s^2),
and t is the time interval (12 s).

Plugging in the values, we have:

s = (0)(12) + (1/2)(6.2)(12)^2

Simplifying further:

s = 0 + (1/2)(6.2)(144)

s = 0 + (3.1)(144)

s = 0 + 446.4

Therefore, the approximate distance traveled by the airplane during the 12 second interval is approximately 446.4 meters.

To find the distance traveled by the airplane during this 12-second interval, we can use the equation:

distance = initial velocity * time + 0.5 * acceleration * time^2

Since the airplane was originally at rest (initial velocity = 0), the equation simplifies to:

distance = 0.5 * acceleration * time^2

Plugging in the values, we get:

distance = 0.5 * 6.2 m/s^2 * (12 s)^2

Simplifying the equation further:

distance = 0.5 * 6.2 m/s^2 * 144 s^2

distance = 442.8 m

Therefore, the approximate distance traveled by the airplane during this 12-second interval is approximately 442.8 meters.

s=at^2/2