A helicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. Determine the acceleration of the helicopter and its initial velocity.
Assuming it is uniform motion,then
Vf^2=Vi^2+2ad
a=-Vi^2/1200
and
vf=vi+at
0=Vi-Vi^2/1200 *5*60
0=Vi(1-Vi/300)
Vi=300m/s
a=..
check my work.
Notice Vi is positive, it was going up at a rapid rate, if these calcs are correct.
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To determine the acceleration and initial velocity of the helicopter, we can use the equations of motion:
1. Displacement equation: s = ut + (1/2)at^2
2. Final velocity equation: v = u + at
3. Average velocity equation: v_avg = (u + v)/2
Given:
- Initial height, s = 600 m
- Time taken, t = 5.00 minutes = 5.00 * 60 = 300 seconds
- Final velocity, v = 0 m/s (reaches the ground at rest)
First, let's calculate the acceleration (a):
From equation 1, we can rewrite it as:
s = (1/2)at^2
Rearranging the equation:
a = (2s) / t^2
Substituting the given values:
a = (2 * 600 m) / (300 s)^2
a = 0.267 m/s^2
Therefore, the acceleration of the helicopter is 0.267 m/s^2.
Next, let's calculate the initial velocity (u):
From equation 2:
v = u + at
Rearranging the equation:
u = v - at
Substituting the given values:
u = 0 m/s - (0.267 m/s^2)(300 s)
u = -80.1 m/s
Therefore, the initial velocity of the helicopter is -80.1 m/s. The negative sign indicates that the helicopter is descending.