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.
To find the magnitude and direction of the jet's acceleration, we can use the equations of motion.
The equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d) is given by:
vf^2 = vi^2 + 2ad
Given:
Initial velocity (vi) = 78.3 m/s
Displacement (d) = 959 m
Final velocity (vf) = 0 (since the jet comes to rest)
Substituting the given values into the equation, we have:
0 = (78.3 m/s)^2 + 2a(959 m)
Simplifying the equation, we get:
0 = 6142.49 m^2/s^2 + 1918a
Solving for acceleration (a), we have:
a = - 6142.49 m^2/s^2 / 1918
Calculating the value of a, we get:
a ≈ -3.201 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the motion, which means it is pointing north (opposite to the south direction).
Therefore, the magnitude of the jet's acceleration is approximately 3.201 m/s^2, and its direction is north.
To find the magnitude and direction of the jet's acceleration, we can use the equations of motion.
First, let's assign the given values:
Initial velocity (u) = 78.3 m/s (south)
Final velocity (v) = 0 m/s (jet comes to rest)
Displacement (s) = 959 m (south)
Acceleration (a) = ? (magnitude and direction)
We'll use the equation:
v^2 = u^2 + 2as
Rearranging the equation to solve for 'a', we get:
a = (v^2 - u^2) / (2s)
Substituting the given values:
a = (0^2 - 78.3^2) / (2 * 959)
Calculating the acceleration:
a = (-6138.69) / (1918)
a ≈ -3.2 m/s^2
The negative sign indicates that the jet decelerates or slows down in the direction of its initial motion.
Therefore, the magnitude of the acceleration is approximately 3.2 m/s^2, and its direction is due south (opposite to the initial motion of the jet).
a=[v²(fin)-v²(init)]/2•s= - 78.3²/2•959 = = -3.2 m/s².
"a "is opposite to the velocity, i.e., to the north.