Landing with a speed of 78.3 m/s, and traveling due south, a jet comes to rest in 959 m. Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.
3.2 m/s^2
initial velocity = 78.3 m/s
final velocity = 0 m/s
distance traveled - 959 m
a = ?
use
(final velocity)^2 = (initial velocity)^2 + 2ad
solve equation for 'a'.
3.2
omg
To find the magnitude and direction of the jet's acceleration, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the jet comes to rest)
u = initial velocity (78.3 m/s)
a = acceleration (unknown)
s = distance traveled (959 m)
Rearranging the equation to solve for acceleration (a), we have:
a = (v^2 - u^2) / (2s)
Substituting the given values into the equation:
a = (0 - (78.3)^2) / (2 * 959)
a = (-6112.89) / (1918)
a ≈ -3.19 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (i.e., north).
Therefore, the magnitude of the jet's acceleration is approximately 3.19 m/s^2, and its direction is north.